# How do you solve x^2 - 3x - 4 = 0 using the quadratic formula?

May 19, 2016

$x = - 1$
$x = 4$

#### Explanation:

The quadratic formula is $\frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$.

In a trinomial written in the form $a {x}^{2} + b x + c$,
$a$- the coefficient (number) in front of the ${x}^{2}$
$b$- the coefficient in front of the $x$
$c$- the constant (number by itself and has no variable- in this case, has no $x$- attached to it

So, in this problem,
$a = 1$ (no coefficient is an invisible $1$)
$b = - 3$ (don't forget about the negative!)
$c = - 4$

Now, we plug all these values into the quadratic formula with the corresponding values:
$\frac{- - 3 \pm \sqrt{{\left(- 3\right)}^{2} - 4 \left(1\right) \left(- 4\right)}}{2 \left(1\right)}$
$\frac{3 \pm \sqrt{9 - 4 \left(- 4\right)}}{2}$
$\frac{3 \pm \sqrt{9 + 16}}{2}$
$\frac{3 \pm \sqrt{25}}{2}$
$\frac{3 \pm 5}{2}$

Now, you have to solve for BOTH solutions:

One solution: $x =$ $\frac{3 + 5}{2}$ = $\frac{8}{2}$ = $4$

Second solution: $x =$ $\frac{3 - 5}{2}$ = $- \frac{2}{2}$ = $- 1$

So, the two solutions are $x = - 1$ and $x = 4$