# How do you solve 4x^2+9x+5=0 using the quadratic formula?

May 13, 2016

$\text{ "x= -10/8" and } - 1$

#### Explanation:

Given that the standard form equation is: $y = a {x}^{2} + b x + c$

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

Given equation:$\text{ } y = 4 {x}^{2} + 9 x + 5$

So:$\text{ "a=4" ; "b=9" ; } c = 5$

Thus by substitution:

$\text{ } x = \frac{- 9 \pm \sqrt{{9}^{2} - 4 \left(4\right) \left(5\right)}}{2 \left(4\right)}$

$\text{ } x = \frac{- 9 \pm \sqrt{81 - 80}}{8}$

$\text{ } x = \frac{- 9 \pm 1}{8}$

$\text{ "x= -10/8" and } - 1$