The quadratic formula is one tool that can be used to find the roots of a quadratic equation.  It is written:

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

where the values of $a$, $b$, and $c$ are the values of the coefficients in the quadratic equation:

$ax^2+bx+c=0$

The quadratic formula can always be used to find the roots of a quadratic equation, regardless of whether the roots are real or complex, whole numbers or fractions, and so on.

Criteria For Use

To use the quadratic formula, two criteria must be satisfied:

1. The quadratic equation must equal zero;  $ax^2+bx+c=0$
2. $a$ must not equal zero

The first criterion must be satisfied to use the quadratic formula because conceptually, the formula gives the values of $x$ where the quadratic function $f(x) = ax^2+bx+c = 0$; the roots of the quadratic function.

You can see why the second condition must be true by looking at the quadratic formula. If $a=0$, the denominator of the formula is zero, which results in an undefined quantity. Conceptually, this makes sense because if $a=0$, then the function $f(x) = ax^2 + bx+c$ is not quadratic, but linear!

Solving Quadratic Equations with the Quadratic Formula

Solutions to $ax^2 + bx+c =0$ can be found by using the quadratic formula

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

The symbol ± indicates there will be two solutions, one that involves adding the square root of $b^2-4ac$, and the other found by subtracting said square root. The resulting $x$ values (zeros) may or may not be distinct, and may or may not be real.

Example

Let's take a look at an example.  Suppose we want to find the roots of the following quadratic function:

$f(x) = 2x^2+5x+3$

First, we need to set the function equal to zero, as the roots are where the function equals zero.

$0 = 2x^2+5x+3$

Second, we need to identify the constants in the equation.  The value of $a$ is two, the value of $b$ is five, and the value of $c$ is three.  We can now substitute these values into the quadratic equation and simplify:

$x=\dfrac{-5 \pm \sqrt{5^2-4(2)(3)}}{2(2)}$

$x= \dfrac{-5 \pm \sqrt{25-24}}{4}$

$x = \dfrac{-5 \pm \sqrt{1}}{4}$

$\displaystyle x = \frac{-5}{4} + \frac{1}{4}$, $\displaystyle \frac{-5}{4} - \frac{1}{4}$

$x = \dfrac{-3}{4}$, $\dfrac{-6}{4}$

$x=\dfrac{-3}{4}$, $\dfrac{-3}{2}$