Quadratic Equations: Discriminant & How to Solve Them

A quadratic equation is any equation that can be written in the standard form

$$ax^2 + bx + c = 0, \qquad a \neq 0$$

where $a$, $b$, and $c$ are constants and $x$ is the unknown. The condition $a \neq 0$ is what makes it quadratic: if $a = 0$, the $x^2$ term disappears and you are left with a linear equation. Because the highest power of $x$ is two, a quadratic equation has at most two real solutions, called its roots. This guide covers the three main solving methods — the quadratic formula, factoring, and completing the square — along with the discriminant, which tells you how many real roots to expect before you solve.

Standard Form And What The Coefficients Mean

Putting an equation in standard form is almost always the first step, because every method below assumes one side equals zero. For example, $2x^2 + 5x = 3$ becomes $2x^2 + 5x - 3 = 0$, so $a = 2$, $b = 5$, and $c = -3$. Identifying $a$, $b$, $c$ correctly — including their signs — is the single most important setup step. A misplaced sign here propagates through every later calculation.

The Discriminant: How Many Real Roots?

The discriminant is the quantity under the square root in the quadratic formula:

$$\Delta = b^2 - 4ac$$

It tells you the nature of the roots without solving the whole equation:

Discriminant         Roots
------------------   ------------------------------------------
Δ > 0                Two distinct real roots
Δ = 0                One repeated real root (a double root)
Δ < 0                No real roots (two complex conjugate roots)

For $2x^2 + 5x - 3 = 0$, the discriminant is

$$\Delta = 5^2 - 4(2)(-3) = 25 + 24 = 49$$

Since $49 > 0$, there are two distinct real roots. And because $49 = 7^2$ is a perfect square, the roots will be rational, which is a hint that factoring may work cleanly.

The Quadratic Formula

The quadratic formula solves any quadratic equation in standard form:

$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

The $\pm$ produces the two roots. This is the method to reach for when factoring is not obvious, since it never fails as long as $a \neq 0$.

Why The Formula Works (Derivation)

The quadratic formula is not magic — it is completing the square done once, in general. Start from standard form, divide by $a$, move the constant to the right, and add $\left(\frac{b}{2a}\right)^2$ to both sides. The left side becomes a perfect square, and combining the right side over a common denominator gives

$$\left(x + \frac{b}{2a}\right)^2 = \frac{b^2 - 4ac}{4a^2}$$

Take the square root of both sides (keeping the $\pm$) and isolate $x$:

$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

This derivation also explains why the discriminant $b^2 - 4ac$ controls the number of roots: it is the quantity sitting under the square root.

Solving By Factoring

Factoring is the fastest method when the quadratic factors nicely over the integers. The idea relies on the zero-product property: if a product equals zero, at least one factor is zero.

To factor $2x^2 + 5x - 3 = 0$, look for two numbers that multiply to $a \cdot c = (2)(-3) = -6$ and add to $b = 5$. Those numbers are $6$ and $-1$. Split the middle term and factor by grouping:

$$2x^2 + 6x - x - 3 = 0 \;\Rightarrow\; 2x(x + 3) - 1(x + 3) = 0 \;\Rightarrow\; (2x - 1)(x + 3) = 0$$

Setting each factor to zero gives

$$2x - 1 = 0 \;\Rightarrow\; x = \tfrac{1}{2}, \qquad x + 3 = 0 \;\Rightarrow\; x = -3$$

Solving By Completing The Square

Completing the square works for every quadratic and is the method behind the formula. Take $x^2 + 6x - 7 = 0$, move the constant across, and add $\left(\frac{6}{2}\right)^2 = 9$ to both sides to form a perfect square:

$$x^2 + 6x + 9 = 16 \;\Rightarrow\; (x + 3)^2 = 16$$

Take square roots and solve:

$$x + 3 = \pm 4 \;\Rightarrow\; x = 1 \;\text{ or }\; x = -7$$

This method is especially useful for finding the vertex of a parabola, since it rewrites the quadratic in vertex form.

Choosing A Method

Situation                              Best method
------------------------------------   --------------------------
Factors over the integers              Factoring (fastest)
Messy / non-integer coefficients       Quadratic formula
Need vertex form or a proof            Completing the square
Just need root count                   Discriminant only

Worked Example 1: Quadratic Formula

Solve $3x^2 - 5x - 2 = 0$.

Here $a = 3$, $b = -5$, $c = -2$. Compute the discriminant first:

$$\Delta = (-5)^2 - 4(3)(-2) = 25 + 24 = 49$$

Since $\Delta = 49 > 0$, expect two real roots. Apply the formula:

$$x = \frac{-(-5) \pm \sqrt{49}}{2(3)} = \frac{5 \pm 7}{6}$$ $$x = \frac{12}{6} = 2 \qquad \text{or} \qquad x = \frac{-2}{6} = -\frac{1}{3}$$

So the roots are $x = 2$ and $x = -\tfrac{1}{3}$.

Worked Example 2: A Repeated Root

Solve $x^2 - 6x + 9 = 0$.

With $a = 1$, $b = -6$, $c = 9$, the discriminant is

$$\Delta = (-6)^2 - 4(1)(9) = 36 - 36 = 0$$

Because $\Delta = 0$, there is exactly one repeated root. The formula gives

$$x = \frac{6 \pm \sqrt{0}}{2} = 3$$

This matches factoring, since $x^2 - 6x + 9 = (x - 3)^2$, confirming the double root at $x = 3$. Here the parabola just touches the $x$-axis at one point instead of crossing it twice.

Now You Try

Solve $2x^2 - 7x + 3 = 0$ by two different methods and check that they agree. First compute the discriminant ($\Delta = 49 - 24 = 25$), so two rational roots exist; then try factoring by finding numbers that multiply to $a\cdot c = 6$ and add to $-7$. Confirm with the quadratic formula. Your roots should be $x = 3$ and $x = \tfrac{1}{2}$ — if both methods land on the same pair, you have verified your work.

Calculation Traps To Watch

Sign errors in $b^2 - 4ac$ are the most common mistake; when $c$ is negative, $-4ac$ becomes positive, so $\Delta$ grows rather than shrinks. The second trap is forgetting that the whole numerator $-b$ is divided by $2a$, not just the square-root part — write the fraction bar across the entire numerator. Third, students sometimes divide both sides by $x$ to "simplify," which silently discards the root $x = 0$; factor instead. Finally, never drop the $\pm$: a quadratic almost always has two roots, and keeping only the plus sign loses one of them.

Where Quadratic Equations Show Up

Quadratics model projectile motion, area and optimization problems, and the parabolic shapes that appear throughout physics and engineering. Mastering the standard form, the discriminant, and all three solving methods gives you a complete toolkit: the discriminant tells you what to expect, factoring handles the clean cases quickly, and the quadratic formula closes out everything else.