# How do you find the roots, real and imaginary, of y= 5x^2 - 6x - (2x+1)^2  using the quadratic formula?

Oct 20, 2017

Two Real roots at $x = 5 \pm \sqrt{26}$

#### Explanation:

$y = 5 {x}^{2} - 6 x - {\left(2 x + 1\right)}^{2}$

$y = 5 {x}^{2} - 6 x - \left(4 {x}^{2} + 4 x + 1\right)$

$\rightarrow y = \textcolor{red}{1} {x}^{2} \textcolor{b l u e}{- 10} x \textcolor{g r e e n}{- 1}$

Applying the quadratic formula for roots:
$\textcolor{w h i t e}{\text{XXX}} x = \frac{- \textcolor{b l u e}{b} \pm \sqrt{{\textcolor{b l u e}{b}}^{2} - 4 \textcolor{red}{a} \textcolor{g r e e n}{c}}}{2 \textcolor{red}{a}}$
we have
$\textcolor{w h i t e}{\text{XXX}} x = \frac{- \textcolor{b l u e}{\left(- 10\right)} \pm \sqrt{{\textcolor{b l u e}{\left(- 10\right)}}^{2} - 4 \times \textcolor{red}{1} \times \textcolor{g r e e n}{\left(- 1\right)}}}{2 \times \textcolor{red}{1}}$

$\textcolor{w h i t e}{\text{XXX}} x = \frac{10 \pm \sqrt{100 + 4}}{2}$

$\textcolor{w h i t e}{\text{XXX}} x = 5 \pm \frac{\sqrt{104}}{2}$

$\textcolor{w h i t e}{\text{XXX}} x = 5 \pm \sqrt{26}$