Forms of Quadratic Equations: Standard, Vertex and Factored (With Examples)

What's the quadratic equation? Definition in plain English

The defining feature: a squared term and nothing higher

A quadratic equation is any equation that can be written with a squared variable term, a linear term, and a constant, with no higher power anywhere. In symbols: $ax^2 + bx + c = 0$, where $a \neq 0$. Drop the squared term and it's linear; add a cubed term and it's no longer quadratic.

Quadratic equation vs quadratic formula vs quadratic function

Three related terms get fused together in student searches, and they aren't the same thing. The quadratic equation is the statement $ax^2+bx+c=0$ you're trying to solve. The quadratic formula is one specific tool for solving it: $x = \dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$. The quadratic function is the related graphing object, $y = ax^2+bx+c$, whose graph is a parabola. Solving the equation means finding where that parabola crosses the x-axis.

Quadratic equation def in one sentence

Quadratic equation def, short version: an equation where the highest power of the variable is 2, written in one of three interchangeable forms of the quadratic equation — standard, vertex, or factored — each of which hands you a different piece of information about the parabola without any extra work. This shows up first in Algebra 1 as an introduction, returns in Algebra 2 as a full unit, and appears directly on SAT Math and under Common Core standard HSA-REI.B.4.

The three forms, side by side

The forms of a quadratic equation are just different ways to write the same relationship — every one describes the exact same parabola. The differences are only in what each one hands you immediately, and what it forces you to calculate. Take one concrete example and hold it constant: $y = x^2 + 6x + 5$.

Standard form: ax^2 + bx + c - reads off the y-intercept

Standard form is $y = x^2 + 6x + 5$, with $a=1$, $b=6$, $c=5$. Read the y-intercept straight off $c$: the parabola crosses the y-axis at $(0, 5)$. Standard form is also the form the quadratic formula expects, so it's the starting point for solving.

Vertex form: a(x - h)^2 + k - reads off the vertex and axis of symmetry

The same parabola in vertex form is $y = (x+3)^2 - 4$. The vertex sits at $(h, k) = (-3, -4)$, and the axis of symmetry is the vertical line $x = -3$. Vertex form is the fastest form for reading a maximum or minimum value, which is why word problems that ask "what is the highest point" almost always want this form.

Factored form: a(x - r)(x - s) - reads off the roots

The same parabola in factored form is $y = (x+1)(x+5)$. The roots — where the parabola crosses the x-axis — are $x=-1$ and $x=-5$, read directly from the two factors. Factored form is the fastest form for solving the equation, when the numbers cooperate.

One parabola, three ways to write it

This is the same parabola every one of the different forms of quadratic equations describes — the equation just gets rearranged, not changed.

Convert between forms without a calculator

Once you can move fluently between the forms of the quadratic equation, factoring and vertex questions stop feeling like separate topics — they're the same six or seven lines of algebra run in different directions.

Standard to vertex: completing the square, six lines

Completing the square turns standard form into vertex form. Starting from $y = x^2+6x+5$:

1. Isolate the variable terms: $x^2+6x = y - 5$.
2. Take half of $b$ ($6 \div 2 = 3$) and square it ($3^2=9$).
3. Add 9 to both sides: $x^2+6x+9 = y-5+9$.
4. The left side is now a perfect square: $(x+3)^2 = y+4$.
5. Solve for $y$: $y = (x+3)^2 - 4$.
6. Read off the vertex: $(-3,-4)$.

Standard to factored: factoring and the AC method

For $x^2+6x+5$, find two numbers that multiply to $c=5$ and add to $b=6$: that's 1 and 5. So $x^2+6x+5=(x+1)(x+5)$. When $a \neq 1$, use the AC method: multiply $a \times c$, find two numbers that multiply to that product and add to $b$, split the middle term, then factor by grouping.

Vertex to standard: expand and collect

From $y=(x+3)^2-4$: expand the square, $(x+3)^2 = x^2+6x+9$, then collect: $y = x^2+6x+9-4 = x^2+6x+5$.

Factored to standard: FOIL and collect

From $y=(x+1)(x+5)$: FOIL gives $x^2+5x+x+5$, and collecting like terms gives $y=x^2+6x+5$.

Which conversion is worth doing on a timed test

Not every conversion is worth the minutes. If the question only asks for the y-intercept, you don't need vertex form at all — standard form already has it. If the question asks for the vertex and the equation is already in standard form with a friendly $b$, complete the square; don't factor first and then guess the midpoint. If the question asks for the roots and the numbers are ugly, skip conversion entirely and go straight to the quadratic formula.

Quadratic equation examples in each form

These examples of quadratic equations cover every combination a test writer reaches for: $a=1$ vs $a\neq1$, a positive vs a negative leading coefficient, and a repeated root. Each is an algebraic equation example you can check yourself by substitution.

Standard-form examples (including a = 1 and a != 1)

• $y = x^2 - 5x + 6$ ($a=1$): y-intercept $(0,6)$.
• $y = 3x^2 - 2x - 1$ ($a \neq 1$): y-intercept $(0,-1)$.
• $y = x^2 + 8x + 16$ ($a=1$, a perfect-square trinomial): y-intercept $(0,16)$.

Vertex-form examples (including a negative a)

• $y = (x-2)^2 + 3$: vertex $(2,3)$, opens upward.
• $y = -2(x+1)^2 + 5$ (negative $a$): vertex $(-1,5)$, opens downward — the vertex is a maximum, not a minimum.
• $y = 3(x-4)^2 - 1$: vertex $(4,-1)$.

Factored-form examples (including a repeated root)

• $y = (x-2)(x-5)$: roots $x=2, x=5$.
• $y = 2(x+3)(x-1)$: roots $x=-3, x=1$.
• $y = (x-4)^2$ (a repeated root): the parabola touches the x-axis once, at $x=4$, instead of crossing it.

A random quadratic equation, worked cold

Take a random quadratic equation with no special setup: $y = -3x^2+5x+2$. Here $a=-3$, $b=5$, $c=2$. The discriminant is $b^2-4ac = 25-4(-3)(2)=25+24=49$, a perfect square, so it factors cleanly: $x=\dfrac{-5\pm7}{-6}$, giving $x=-\dfrac{1}{3}$ and $x=2$. In factored form, $y=-3\left(x+\dfrac13\right)(x-2)$, or equivalently $y=-(3x+1)(x-2)$.

Solving quadratic equations using the quadratic formula, step by step

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:

$x = \frac{-4 \pm \sqrt{4^2-4(1)(1)}}{2(1)}$

Simplify the discriminant

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

Two roots, one root, or none - what b^2 - 4ac tells you

The discriminant $b^2-4ac$ tells you what kind of roots you're about to get, before you finish solving.

• Positive ($12 > 0$ 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: $x=\dfrac{-4\pm2\sqrt3}{2}=-2\pm\sqrt3$. 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.

When to use the quadratic formula vs factoring vs completing the square

When to use quadratic equation methods like factoring versus the formula comes down to one quick check on the coefficients, not a memorized rule.

Factor first if the numbers are friendly

Try factoring before anything else when $a=1$ and $c$ looks like it splits cleanly — it's the fastest path and needs no formula at all.

Complete the square when the question wants the vertex

If the question explicitly asks for the vertex, the maximum height, or the axis of symmetry, complete the square directly rather than factoring and then hunting for the midpoint of the roots.

Use the formula when nothing factors

The quadratic formula always works, for any $a$, $b$, $c$ — including when the roots are irrational or complex. Use it as the fallback the moment factoring stalls past a few seconds. Using quadratic equation methods interchangeably like this becomes automatic once you've run the triage a few times.

The 15-second triage

Glance at $c$ first. If two integer factors of $c$ obviously add to $b$, factor. If not, and the question wants the vertex, complete the square. Otherwise, go straight to the formula — don't burn a minute testing factor pairs that were never going to work.

Forming a quadratic equation from roots or from a word problem

Forming a quadratic equation runs every conversion above in reverse — starting from what you know about the graph, not from the equation itself.

From two roots: the reverse of factored form

Given roots $x=3$ and $x=-2$, reverse factored form: $(x-3)(x+2)=0$. Expand: $x^2-x-6=0$.

From a vertex and a point

Given vertex $(2,-3)$ and a point $(4,5)$ on the parabola, start from vertex form: $y=a(x-2)^2-3$. Substitute the point: $5=a(4-2)^2-3$, so $5=4a-3$, and $a=2$. The equation is $y=2(x-2)^2-3$, or in standard form, $y=2x^2-8x+5$. Plotting a vertex from a point this way uses the same coordinate-plane skills covered in geometry help.

From a projectile-height word problem

Projectile height follows $h(t) = -16t^2+v_0t+h_0$ (feet and seconds), where $v_0$ is initial velocity and $h_0$ is starting height. A ball thrown from 5 feet with an initial velocity of 48 ft/s gives $h(t)=-16t^2+48t+5$. The vertex of this parabola is the maximum height; the positive root is the moment the ball lands. The same setup reappears with real units and a free-body diagram on physics homework help.

Practice questions for quadratic equations, with answers

The questions for quadratic equation practice below mirror what shows up on a chapter test: identify, convert, then solve. Answers are given directly under each item.

Identify the form (6 items)

1. $y=2x^2-3x+1$ — Answer: standard.
2. $y=(x-1)(x+4)$ — Answer: factored.
3. $y=-(x+2)^2+7$ — Answer: vertex.
4. $y=x^2-9$ — Answer: standard (also factorable as $(x-3)(x+3)$).
5. $y=3(x-5)^2$ — Answer: vertex.
6. $y=(x+6)(x-6)$ — Answer: factored.

Convert the form (4 items)

1. Convert $y=x^2-4x+3$ to factored form. — Answer: $y=(x-1)(x-3)$.
2. Convert $y=x^2-4x+3$ to vertex form. — Answer: $y=(x-2)^2-1$.
3. Convert $y=(x+2)^2-9$ to standard form. — Answer: $y=x^2+4x-5$.
4. Convert $y=(x-3)(x+1)$ to standard form. — Answer: $y=x^2-2x-3$.

Solve by the best method (6 items)

1. $x^2-7x+12=0$ — Answer: factor; $x=3,4$.
2. $x^2+6x+9=0$ — Answer: factor (perfect square); $x=-3$ (repeated root).
3. $2x^2-8=0$ — Answer: isolate and root; $x=\pm2$.
4. $x^2+3x-1=0$ — Answer: formula; $x=\dfrac{-3\pm\sqrt{13}}{2}$.
5. $x^2-4x-1=0$ — Answer: formula; $x=2\pm\sqrt5$.
6. $x^2+x+1=0$ — Answer: formula; discriminant $=1-4=-3$, complex roots $x=\dfrac{-1\pm i\sqrt3}{2}$.

When the quadratics unit is graded and there are forty of them

The forms themselves are learnable in an evening — one example, three rewrites, and the pattern sticks. A problem set with forty conversions and word problems due by Thursday is where most students go looking to do my math homework instead of grinding through every item by hand. GradeDraft's math homework desk solves quadratics problem sets — conversions, the formula, word problems, and the practice-bank style items above — with every step shown, not just a final answer, priced from $25 per problem after a quick look at the set.

For the algebra that comes before quadratics, see solving linear equations step by step and solving systems by elimination. If your course runs the CPM curriculum, CPM Algebra 2 homework help covers the same forms in that textbook's notation. Quadratics also show up outside math class: a chemistry equilibrium problem often reduces to one, covered on chemistry and biology homework help, and the bowed-out shape of an economics production possibilities frontier draws on the same curve, covered on what are economic resources.

Free practice banks like Albert.io, Cuemath, Vaia, Expii, and Media4Math cover similar problem types, and graphing the result in Desmos or GeoGebra is a fast way to check any conversion above by eye.