# How do you find the roots, real and imaginary, of y= x^2-39x-2(3x-1)^2  using the quadratic formula?

Jul 17, 2017

$0.04 \text{ or } 2.644$

#### Explanation:

Right now, you have $y = {x}^{2} - 39 x - 2 {\left(3 x - 1\right)}^{2}$

First, we need to expand ${\left(3 x - 1\right)}^{2}$.

To expand the brackets in the form of ${\left(\textcolor{red}{a} \textcolor{g r e e n}{x} + \textcolor{b l u e}{b}\right)}^{2}$, we do ${\left(\textcolor{red}{a} \textcolor{g r e e n}{x}\right)}^{2} + 2 \left(\textcolor{red}{a} \textcolor{g r e e n}{x} \textcolor{b l u e}{b}\right) + {\textcolor{b l u e}{b}}^{2}$

In this case we have ${\left(\textcolor{red}{3} \textcolor{g r e e n}{x} + \textcolor{b l u e}{- 1}\right)}^{2}$

${\left(3 x\right)}^{2} = 9 {x}^{2}$
$2 \cdot 3 x \cdot - 1 = - 6 x$
$- {1}^{2} = 1$
$9 {x}^{2} - 6 x + 1$

But we need 2 lots, so $2 \left(9 {x}^{2} - 6 x + 1\right) = {18}^{2} - 12 x + 2$

$y = {x}^{2} - 39 x + 18 {x}^{2} - 12 x + 2 = 19 {x}^{2} - 51 x + 2$

The quadratic formula is $x = \frac{- \textcolor{b l u e}{b} \pm \sqrt{{\textcolor{b l u e}{b}}^{2} - 4 \textcolor{red}{a} \textcolor{\mathmr{and} a n \ge}{c}}}{2 \textcolor{red}{a}} \equiv \frac{- \textcolor{b l u e}{- 51} \pm \sqrt{{\textcolor{b l u e}{- 51}}^{2} - 4 \left(\textcolor{red}{19} \cdot \textcolor{\mathmr{and} a n \ge}{4}\right)}}{2 \left(\textcolor{red}{19}\right)} = \frac{\textcolor{b l u e}{51} \pm \sqrt{\textcolor{b l u e}{2601} - \textcolor{y e l l o w}{152}}}{38} = 0.04 \text{ or } 2.644$