# How do you solve 2x^2+3x+5=0 using the quadratic formula?

Nov 13, 2016

$x = - \frac{3}{4} + \frac{\sqrt{31} i}{4}$ and $x = - \frac{3}{4} - \frac{\sqrt{31} i}{4}$

#### Explanation:

Comparing with the standard form, $a {x}^{2} + b x + c$, we see that:

$a = 2$
$b = 3$
$c = 5$

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

Substitute in the values of a, b, and c:

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

Simplify:

$x = \frac{- 3 \pm \sqrt{- 31}}{4}$

We must factor out $\sqrt{- 1}$ and write it as $i$

$x = - \frac{3}{4} + \frac{\sqrt{31} i}{4}$ and $x = - \frac{3}{4} - \frac{\sqrt{31} i}{4}$