How to Solve Quadratic Equations: Factoring, the Quadratic Formula, Completing the Square and Graphing

§ 01

How to solve a quadratic equation: pick the method first

How to solve a quadratic equation depends on what the numbers actually look like, not on memorizing one universal recipe and forcing every equation through it. Solving quadratic equations quickly means learning which method an equation is asking for before writing anything down — the four methods below all reach the same roots, but three of them are faster than the fourth whenever the numbers cooperate. This page is about the solving side only; if what you actually need is standard, vertex and factored form, compared side by side, that's a separate page, because reading information off an equation and solving it for its roots are two different skills that happen to share the same starting expression.

That triage is how to solve quadratic equations by formula only when the formula is actually the fastest route on offer — glance at $c$ and $b$ first, and reach for the formula last, not first. None of the four methods is more "correct" than the others: every one of them, applied to the same equation, lands on the identical pair of roots, because each method is just a different route through the same underlying algebra. The only thing that changes from method to method is how many lines it takes to get there — and on a timed test, lines saved are the entire point of choosing well.

§ 02

Solving by factoring

a = 1: two numbers that multiply to c and add to b

For $x^2-x-6=0$, find two numbers that multiply to $-6$ and add to $-1$: those are $-3$ and $2$. So $x^2-$$x-$$6=$$x^2-x-6=(x-3)(x+2)$, and the equation is true when either factor is 0: $x=3$ or $x=-2$. Check: $3^2-$$3-$$6=$$9-$$3-$$6=$$3^2-3-6=9-3-6=0$; $(-2{)}^2-$$(-2)-$$6=$$4+$$2-$$6=$$(-2)^2-(-2)-6=4+2-6=0$.

a ≠ 1: the AC method

For $2x^2+7x+3=0$, multiply $a\times c=2\times3=6$, then find two numbers that multiply to $6$ and add to $b=7$: those are $6$ and $1$. Split the middle term and factor by grouping: $2x^2+$$6x+$$x+$$3=$$2x^2+6x+x+3=0$, so $2x(x+3)+$$1(x+3)=$$2x(x+3)+1(x+3)=0$, giving $(2x+1)(x+3)=$$(2x+1)(x+3)=0$. The roots are $x=-\dfrac12$ and $x=-3$. Check: $2(0.25)+$$7(-0.5)+$$3=$$0.5-$$3.5+$$3=$$2(0.25)+7(-0.5)+3=0.5-3.5+3=0$; $2(9)+$$7(-3)+$$3=$$18-$$21+$$3=$$2(9)+7(-3)+3=18-21+3=0$.

Both examples work because of the same underlying idea: multiplying two binomials always produces a middle term that's a sum, so factoring is really just running FOIL backward. The AC method exists only to handle the case where $a\neq1$ makes that sum harder to spot by eye — split the middle term using the pair you found, then group and factor each half, and the same binomial appears in both groups every time the numbers were chosen correctly.

§ 03

The quadratic formula, step by step

What is the quadratic formula?

The quadratic formula solves any quadratic equation written in standard form, $ax^2+bx+c=0$:

It always works, even when factoring stalls or the roots turn out to be irrational or complex — the one method guaranteed to work regardless of what $a$, $b$ and $c$ turn out to be, which is exactly why it's the fallback rather than the first move in the table above.

How do you solve quadratic equations using the quadratic formula? Five moves, applied to the same equation every time.

Identify a, b and c (sign errors start here)

For $x^2+4x+1=0$: $a=1$, $b=4$, $c=1$. The most common error at this step is dropping a negative sign when $b$ or $c$ is negative — write them down before substituting, don't carry them in your head.

Substitute into the formula

Solving quadratic equations using the quadratic formula starts with a direct substitution:

Simplify the discriminant

$4^2-$$4(1)(1)=$$16-$$4=$$4^2-4(1)(1) = 16-4=12$. So $x = \dfrac{-4\pm\sqrt{12}}{2}$.

Two roots, one root, or none: what b² − 4ac tells you

The discriminant $b^2-4ac$ tells you what kind of roots you're about to get, before you finish solving — it's worth computing on its own, as a quick preview, before running the rest of the formula, especially on a timed test where a negative result means the remaining arithmetic will involve an imaginary number.

• Positive ($b^2-4ac=12>0$ above): two distinct real roots.
• Zero: exactly one real root (a repeated root, the vertex touches the x-axis).
• Negative: no real roots — two complex roots instead. For $x^2+2x+5=0$, the discriminant is $4-20=-16$, so the roots are $x=-1\pm2i$.

A full worked solve with irrational roots

Continuing $x=\dfrac{-4\pm\sqrt{12}}{2}$: simplify the radical first, $\sqrt{12}=2\sqrt{3}$, then reduce: $x=$${\displaystyle \frac{-4\pm 2\sqrt{3}}{2}}=$$-2\pm$$x=\dfrac{-4\pm2\sqrt3}{2}=-2\pm\sqrt3$. Leave the answer as $x=-2+\sqrt3$ and $x=-2-\sqrt3$ unless the question asks for decimals — the surd form is exact; $-2+\sqrt3 \approx -0.27$ and $-2-\sqrt3\approx-3.73$ are rounded.

Notice that $x^2+4x+1=0$ doesn't factor over the integers — there's no pair of whole numbers that multiplies to 1 and adds to 4 — which is exactly the signal from the method table above to reach for the formula instead of hunting for factors that don't exist. The formula never needs that judgment call; it produces the same two roots whether an equation factors neatly or not, which is the whole reason it's the guaranteed fallback.

§ 04

Completing the square to solve

Completing the square solves a quadratic without factoring and without the formula, by rebuilding one side into a perfect square. Take $x^2-2x-15=0$:

1. Isolate the variable terms: $x^2-2x=15$.
2. Take half of $b$ ($-2\div2=-1$) and square it ($(-1)^2=1$).
3. Add 1 to both sides: $x^2-$$2x+$$1=$$15+$$1=$$x^2-2x+1=15+1=16$.
4. The left side is now a perfect square: $(x-1)^2=16$.
5. Square root both sides: $x-1=\pm4$.
6. Solve: $x=1\pm4$, giving $x=5$ or $x=-3$.

Check: $5^2-$$2(5)-$$15=$$25-$$10-$$15=$$5^2-2(5)-15=25-10-15=0$; $(-3{)}^2-$$2(-3)-$$15=$$9+$$6-$$15=$$(-3)^2-2(-3)-15=9+6-15=0$. This is the same move used to convert standard form into vertex form: step 4 is the vertex form $y=(x-1)^2-16$ set equal to zero, and solving continues two lines further, to the roots. Completing the square works on any quadratic, not just this one, because adding $(b/2)^2$ to a $x^2+bx$ expression is always exactly the amount needed to turn it into a perfect square trinomial — that's not a coincidence specific to this equation, it's the algebraic identity the whole method is built on.

§ 05

Solving by square roots and by graphing

x² = k: isolate and root

When an equation has no $x$ term at all, isolate $x^2$ and take the square root of both sides — remembering the $\pm$, since both a positive and a negative number square to the same result. For $3x^2-27=0$: add 27, $3x^2=27$; divide by 3, $x^2=9$; root both sides, $x=\pm3$. Check: $3(9)-$$27=$$27-$$27=$$3(9)-27=27-27=0$. Forgetting the $\pm$ is the single most common error on this method — a calculator's square-root key only ever returns the positive root, so the negative one has to be added back in by hand.

Reading roots from a graph

A graphed parabola crosses the x-axis exactly at its roots, so once the curve is drawn accurately, the roots can be read off directly without any algebra. For $y=x^2-4$, a table of points shows where the curve crosses zero:

Scroll the table sideways to see every column.

The curve crosses $y=0$ at $x=-2$ and $x=2$ — the same two roots factoring gives for $x^2-$$4=$$x^2-4=(x-2)(x+2)$. Graphing is slower to set up than the other three methods, but it's the only one that shows the whole shape of the parabola at once, not just its roots: where the vertex sits, which way the parabola opens, and how close together or far apart the two roots are. A parabola that only touches the x-axis once, without crossing it, is the graphical picture of a repeated root — the same case example 2 below reaches algebraically.

§ 06

Quadratic equation examples, solved

These quadratic equation examples run through every method above on fresh numbers, so you can match a new problem to the method it needs on sight. One of the examples below (5) needs the quadratic formula outright and one (7) has no real roots; the rest factor, root out directly, or reduce to one repeated root.

1. $x^2+2x-8=0$ — factors: $(x+4)(x-2)=$$(x+4)(x-2)=0$, so $x=-4,2$. Check: $16-8-8=0$; $4+4-8=0$.
2. $x^2-10x+25=0$ — a perfect square: $(x-5)^2=0$, so $x=5$ (a repeated root). Check: $25-50+25=0$.
3. $2x^2-3x-2=0$ — AC method, $a\times c=-4$, numbers $-4$ and $1$: $2x(x-2)+$$1(x-2)=$$(2x+1)(x-2)=$$2x(x-2)+1(x-2)=(2x+1)(x-2)=0$, so $x=-\dfrac12,2$. Check: $2(0.25)-$$3(-0.5)-$$2=$$0.5+$$1.5-$$2=$$2(0.25)-3(-0.5)-2=0.5+1.5-2=0$; $2(4)-6-2=0$.
4. $x^2+5x=0$ — factor out $x$: $x(x+5)=0$, so $x=0,-5$. Check: $0+0=0$; $25-25=0$.
5. $x^2-2x-1=0$ — formula: discriminant $=4+4=8$, so $x=\dfrac{2\pm\sqrt8}{2}=1\pm\sqrt2$. Check (decimal): $1+\sqrt2\approx2.414$; $(2.414{)}^2-$$2(2.414)-$$1\approx$$(2.414)^2-2(2.414)-1\approx0$.
6. $-x^2+4x-3=0$ — multiply by $-1$ first: $x^2-4x+3=0$, so $(x-1)(x-3)=$$(x-1)(x-3)=0$, giving $x=1,3$. Check in the original: $-1+4-3=0$; $-9+12-3=0$.
7. $x^2+4=0$ — no real solution: $x^2=-4$, so $x=\pm2i$. The discriminant is $0-16=-16$, negative, which flags the complex roots before any solving happens.
8. $4x^2-1=0$ — square roots: $x^2=\dfrac14$, so $x=\pm\dfrac12$. Check: $4(0.25)-$$1=$$1-$$1=$$4(0.25)-1=1-1=0$.

Read back through the eight: four factor over the integers (1, 2, 4, 6), one needs the AC method (3), one is genuinely irrational (5), one has no real solution (7), and one roots out directly with no factoring at all (8). That spread is deliberate — a real problem set rarely sticks to one method for every item, and recognizing which case an equation belongs to, before picking up a pencil, is most of the actual skill this page teaches, more than any single method's mechanics.

§ 07

Word problems that become quadratics

A word problem becomes a quadratic the moment two quantities get multiplied together and one of them depends on the other — length times width, or velocity and time both feeding into a height formula. The algebra afterward is identical to every worked example above; the only new step is translating the sentence into an equation in the first place.

Projectile height: when does it land?

A ball is thrown upward from a 6-foot platform with an initial velocity of 32 ft/s: $h(t)=$$-16t^2+$$32t+$$h(t)=-16t^2+32t+6$ (feet and seconds). The $-16$ coefficient isn't arbitrary — it comes from half of Earth's gravitational acceleration in feet per second squared, so it shows up in every projectile-height formula written in feet, not just this one. It lands when $h(t)=0$: $-16t^2+$$32t+$$6=$$-16t^2+32t+6=0$. By the quadratic formula, $t=$${\displaystyle \frac{-32\pm \sqrt{{32}^2-4(-16)(6)}}{2(-16)}}=$${\displaystyle \frac{-32\pm \sqrt{1408}}{-32}}=$$1\mp$$t=\dfrac{-32\pm\sqrt{32^2-4(-16)(6)}}{2(-16)}=\dfrac{-32\pm\sqrt{1408}}{-32}=1\mp\dfrac{\sqrt{22}}{4}$. The negative-time root isn't physical — time can't run backward — so it's discarded on sight rather than reported as a second answer: $t=$$1+$${\displaystyle \frac{\sqrt{22}}{4}}\approx$$t=1+\dfrac{\sqrt{22}}{4}\approx2.17$ seconds.

Area: find the missing dimension

A rectangular garden's length is 3 feet more than its width, and its area is 70 square feet. Let the width be $w$, so the length is $w+3$: $w(w+3)=70$, giving $w^2+3w-70=0$. Factoring: $(w+10)(w-7)=$$(w+10)(w-7)=0$, so $w=-10$ (rejected — a width can't be negative) or $w=7$. The width is 7 feet and the length is 10 feet. Check: $7\times10=70$. Rejecting the negative root here works the same way the negative-time root got rejected above: the algebra hands back two mathematically valid solutions, and it's the word problem's own context — a physical width, a moment in time — that rules one of them out, not anything wrong with the quadratic formula or the factoring itself.

Both word problems above turn a sentence into a quadratic and then solve it, which is one direction of a wider pattern worth noticing: many relationships that start out linear turn quadratic the moment area, or the square of a variable, enters the picture. Compare that with direct and quadratic variation, where the same question — does one quantity depend on the square of another — gets asked about tables, graphs and equations instead of word problems.

§ 08

Check the roots by substitution

Checking a quadratic's roots uses the identical method as checking any other equation's solution: substitute the value back into the original equation and confirm both sides match, which is checking a root by substitution applied to a squared variable instead of a linear one. Example 6 above did exactly this in the original, unmultiplied equation, $-x^2+4x-3=0$, rather than the flipped version used to factor it — substituting into the original is what actually confirms the answer, since a sign error in the flip would otherwise go undetected. Every worked example on this page ends the same way, with a substitution check, for the same reason a solving method by itself only tells you an answer is plausible; the check is what actually tells you it's right, not merely reasonable-looking. Decimal roots, like example 5's $1+\sqrt2\approx2.414$, only check approximately once rounded — the exact surd form checks exactly, which is why it's kept as the primary answer and the decimal is offered only as a sanity check.

§ 09

When the quadratics set is due

Four methods are learnable in an evening once you know which one a given equation is asking for — the triage table near the top of this page, applied a dozen times, is usually enough to make the choice automatic. A problem set with thirty quadratics — mixed methods, word problems, and a few with complex roots — due by tomorrow is a different kind of task, and it's where most students go looking for a math set worked with every step shown instead of grinding through every item by hand. GradeDraft's math desk, part of our homework help across math and science, solves sets like this with every step shown, not just a final boxed answer, priced from $25 per problem after a quick look at the set.