# How do you find the roots, real and imaginary, of y= (x+2)^2-x-5  using the quadratic formula?

Jul 18, 2017

See a solution process below:

#### Explanation:

First, we need to expand the squared term on the right side of the equation using the rule:

#(a + b)^2 = a^2 + 2ab + b^2

Substituting $x$ for $a$ and $2$ for $b$ gives:

$y = {\left(x + 2\right)}^{2} - x - 5$

$y = {x}^{2} + \left(2 \times x \times 2\right) + {2}^{2} - x - 5$

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

We can now group and combine like terms to put the equation in standard form for a quadratic.

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

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

$y = {x}^{2} + \left(4 - 1\right) x + \left(4 - 5\right)$

$y = {x}^{2} + 3 x + \left(- 1\right)$

$y = {x}^{2} + 3 x - 1$

For $a {x}^{2} + b x + c = 0$, the values of $x$ which are the solutions to the equation are given by:

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

Substituting $1$ for $a$; $3$ for $b$ and $- 1$ for $c$ gives:

$x = \frac{- 3 \pm \sqrt{{3}^{2} - \left(4 \cdot 1 \cdot - 1\right)}}{2 \cdot 1}$

$x = \frac{- 3 \pm \sqrt{9 - \left(- 4\right)}}{2}$

$x = \frac{- 3 \pm \sqrt{9 + 4}}{2}$

$x = \frac{- 3 \pm \sqrt{13}}{2}$

Or

$x = - \frac{3}{2} + \frac{\sqrt{13}}{2}$ and $x = - \frac{3}{2} - \frac{\sqrt{13}}{2}$