# How do you solve 8x^2 + 4x = -5 using the quadratic formula?

Apr 25, 2017

$x = \frac{- 1 \pm 3 i}{4}$

#### Explanation:

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

where $a {x}^{2} + b x + c = 0$

$8 {x}^{2} + 4 x = - 5$

Move the $- 5$ to the other side of the equation to complete your quadratic equation

$8 {x}^{2} + 4 x + 5 = 0$

Looking at $a {x}^{2} + b x + c = 0$
I can see that your values are...

$a = 8$
$b = 4$
$c = 5$

Now just put those values into the quadratic formula

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

$x = \frac{- \left(4\right) \pm \sqrt{{\left(4\right)}^{2} - 4 \left(8\right) \left(5\right)}}{2 \left(8\right)}$

$x = \frac{- 4 \pm \sqrt{16 - 160}}{16}$

$x = \frac{- 4 \pm \sqrt{- 144}}{16}$

Here we notice that we have a negative square root.
This turns into an imaginary number or $i$ to remove the negative.

$x = \frac{- 4 \pm i \sqrt{144}}{16}$

$x = \frac{- 4 \pm 12 i}{16}$

Notice we have a common factor of $4$ in the numerator and denominator.

$x = \frac{4 \left(- 1 \pm 3 i\right)}{4 \left(4\right)}$

Cancel out common factor

$x = \frac{\cancel{4} \left(- 1 \pm 3 i\right)}{\cancel{4} \left(4\right)}$

$x = \frac{- 1 \pm 3 i}{4}$