# How do you find the roots, real and imaginary, of y=(4x-5)(x+1) using the quadratic formula?

May 16, 2018

${x}_{1} = \frac{5}{4}$
${x}_{2} = - 1$

#### Explanation:

$\left(4 x - 5\right) \cdot \left(x + 1\right)$

$= 4 {x}^{2} + 4 x \cdot 1 - 5 \cdot x - 5$

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

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

${x}_{1 , 2} = \frac{- \left(- 1\right) \pm \sqrt{{\left(- 1\right)}^{2} - 4 \left(4 \left(- 5\right)\right)}}{2 \cdot 4}$

${x}_{1 , 2} = \frac{1 \pm \sqrt{1 + 80}}{8}$

${x}_{1 , 2} = \frac{1 \pm 9}{8}$

${x}_{1} = \frac{5}{4}$

${x}_{2} = - 1$