# How do you find the roots, real and imaginary, of y= x^2 - 3x - (2x+4)^2  using the quadratic formula?

Mar 16, 2016

real roots: $x = - \frac{16}{3} , - 1$
imaginary roots: none

#### Explanation:

$1$ Start by converting the equation into standard form.

$y = {x}^{2} - 3 x - {\left(2 x + 4\right)}^{2}$

y=x^2-3x-(color(crimson)(2x) color(blue)(+4))(color(orange)(2x) color(teal)(+4))

y=x^2-3x-(color(crimson)(2x)(color(orange)(2x)) $\textcolor{c r i m s o n}{+ 2 x} \left(\textcolor{t e a l}{4}\right)$ $\textcolor{b l u e}{+ 4} \left(\textcolor{\mathmr{and} a n \ge}{2 x}\right)$ color(blue)(+4)(color(teal)4))

$y = {x}^{2} - 3 x - \left(4 {x}^{2} + 8 x + 8 x + 16\right)$

$y = {x}^{2} - 3 x - \left(4 {x}^{2} + 16 x + 16\right)$

$y = {x}^{2} - 4 {x}^{2} - 3 x - 16 x - 16$

$y = \textcolor{b r o w n}{- 3} {x}^{2}$ $\textcolor{t u r q u o i s e}{- 19} x$ $\textcolor{v i o \le t}{- 16}$

$2$. Identify the $\textcolor{b r o w n}{a} , \textcolor{t u r q u o i s e}{b} ,$ and $\textcolor{v i o \le t}{c}$ values of the quadratic equation. Then plug the values into the quadratic formula to solve for the roots.

$\textcolor{b r o w n}{a = - 3} \textcolor{w h i t e}{X X X X X} \textcolor{t u r q u o i s e}{b = - 19} \textcolor{w h i t e}{X X X X X} \textcolor{v i o \le t}{c = - 16}$

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

$x = \frac{- \left(\textcolor{t u r q u o i s e}{- 19}\right) \pm \sqrt{{\left(\textcolor{t u r q u o i s e}{- 19}\right)}^{2} - 4 \left(\textcolor{b r o w n}{- 3}\right) \left(\textcolor{v i o \le t}{- 16}\right)}}{2 \left(\textcolor{b r o w n}{- 3}\right)}$

$x = \frac{19 \pm \sqrt{361 - 192}}{-} 6$

$x = \frac{19 \pm \sqrt{169}}{-} 6$

$x = \frac{19 \pm 13}{-} 6$

$x = \frac{19 + 13}{-} 6 \textcolor{w h i t e}{X} , \textcolor{w h i t e}{X} \frac{19 - 13}{-} 6$

$x = \frac{32}{-} 6 \textcolor{w h i t e}{X} , \textcolor{w h i t e}{X} \frac{6}{-} 6$

$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} x = - \frac{16}{3} \textcolor{w h i t e}{i} , - 1 \textcolor{w h i t e}{\frac{a}{a}} |}}}$

$\therefore$, the real roots are $x = - \frac{16}{3}$ or $x = - 1$.