Quadratic formula

When a quadratic will not factor easily, the quadratic formula always works.
For any equation $ax^2 + bx + c = 0$, the solutions are $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.

To use it, first identify $a$, $b$, and $c$ from the equation.
Then substitute those numbers into the formula and simplify carefully, watching the signs.

The $\pm$ sign is what gives a quadratic its two roots — one using $+$ and one using $-$.
The part under the root, $b^2 - 4ac$, is called the discriminant; if it is negative, there are no real solutions.

So if you know one root has the form $\frac{-b + \sqrt{D}}{2a}$, the other must be $\frac{-b - \sqrt{D}}{2a}$.
The two roots differ only in that $\pm$ sign.

Solve $3x^2 + 2x - 4 = 0$ using the quadratic formula.
Here $a = 3$, $b = 2$, $c = -4$, so $x = \frac{-2 \pm \sqrt{2^2 - 4(3)(-4)}}{2(3)}$.
That is $x = \frac{-2 \pm \sqrt{4 + 48}}{6} = \frac{-2 \pm \sqrt{52}}{6}$.

Solve $x^2 - 5x + 6 = 0$ using the quadratic formula.
Here $a = 1$, $b = -5$, $c = 6$, so $x = \frac{5 \pm \sqrt{25 - 24}}{2} = \frac{5 \pm 1}{2}$.
So $x = 3$ or $x = 2$.

A quadratic has one root equal to $\frac{-b + \sqrt{D}}{2a}$. What is the other root?
The two roots differ only in the $\pm$ sign in the formula.
So the other root is $\frac{-b - \sqrt{D}}{2a}$.