# How do you find the roots, real and imaginary, of y= (2x+1)^2-(x + 1) (x - 4)  using the quadratic formula?

Jan 6, 2016

$x = \frac{- 5 + \sqrt{5}}{10}$ or $x = \frac{- 5 - \sqrt{5}}{10}$

#### Explanation:

First expand the equation to get it into standard form.
$y = {\left(2 x + 1\right)}^{2} - \left(x + 1\right) \left(x - 4\right)$
$y = \left(4 {x}^{2} + 2 x + 1\right) - \left({x}^{2} - 3 x - 4\right)$
$y = 3 {x}^{2} + 5 x + 5$
Now use the quadratic formula $x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$ to find the values of $x$ for which $y = 0$
$x = \frac{- 5 \pm \sqrt{{5}^{2} - 4 \cdot 1 \cdot 5}}{2 \cdot 1 \cdot 5}$
$x = \frac{- 5 \pm \sqrt{5}}{10}$
$x = \frac{- 5 + \sqrt{5}}{10}$ or $x = \frac{- 5 - \sqrt{5}}{10}$