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

Oct 15, 2015

$x = \frac{- 5 + 3.87 i}{4} , x = \frac{- 5 - 3.87 i}{4}$

#### Explanation:

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

Where,
a is the constant with x to the power 2.
b is the constant with x to the power 1.
c is the constant.

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

Given Equation:

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

Here,

a = 2
b = 5
c = 5

$x = \frac{- 5 \pm \sqrt{{\left(5\right)}^{2} - 4 \left(2\right) \left(5\right)}}{2 \left(2\right)}$
$x = \frac{- 5 \pm \sqrt{25 - 40}}{4}$
$x = \frac{- 5 \pm \sqrt{- 15}}{4}$
$x = \frac{- 5 \pm 3.87 i}{4}$
$x = \frac{- 5 + 3.87 i}{4} , x = \frac{- 5 - 3.87 i}{4}$