# How do you find the roots, real and imaginary, of y= -3x^2-16x +(x-3)^2+2  using the quadratic formula?

Dec 2, 2017

$2$ real solutions:

$x \setminus \approx - 11.47913 \setminus q \quad , \setminus q \quad x \setminus \approx 0.47913$

#### Explanation:

First let’s simplify the equation by FOILing the ${\left(x - 3\right)}^{2}$ and combining like terms:

${\left(x - 3\right)}^{2} \setminus \quad \setminus \implies \setminus \quad {x}^{2} - 3 x - 3 x + 9 \setminus \quad \setminus \implies \setminus \quad {x}^{2} - 6 x + 9$

Now we can substitute that for ${\left(x - 3\right)}^{2}$ in the original equation:

$\textcolor{b l u e}{- 3 {x}^{2}} - \textcolor{red}{16 x} + \left(\textcolor{b l u e}{{x}^{2}} - \textcolor{red}{6 x} + \textcolor{g r e e n}{9}\right) + \textcolor{g r e e n}{2}$

Combine like terms:

$\textcolor{b l u e}{- 3 {x}^{2}} + \textcolor{b l u e}{{x}^{2}} - \textcolor{red}{16 x} - \textcolor{red}{6 x} + \textcolor{g r e e n}{9} + \textcolor{g r e e n}{2}$

$\setminus \implies \textcolor{b l u e}{- 22 x} - \textcolor{red}{22 x} + \textcolor{g r e e n}{11}$

Now let’s define the variables:

• $\textcolor{b l u e}{a = - 2}$
• color(red)(b=-22
• color(green)(c=11

And plug them into the quadratic formula:

$x = \setminus \frac{\textcolor{red}{- b} \setminus \pm \setminus \sqrt{\textcolor{red}{{b}^{2}} - 4 \left(\textcolor{b l u e}{a}\right) \left(\textcolor{g r e e n}{c}\right)}}{2 \textcolor{b l u e}{a}}$

\implies x=\frac{color(red)(22)\pm\sqrt{(color(red)(-22))-4(color(blue)(-2))(color(green)(11))}}{2(color(blue)(-2))

$\setminus \implies x = \setminus \frac{\textcolor{red}{22} \setminus \pm \setminus \sqrt{\textcolor{red}{484} + 88}}{- 4}$

$\setminus \implies x = \setminus \frac{\textcolor{red}{22} \setminus \pm \setminus \sqrt{572}}{- 4}$

$\setminus \implies x = \setminus \frac{\textcolor{red}{22} \setminus \pm 2 \setminus \sqrt{143}}{- 4}$

$\setminus \implies x = \setminus \frac{11 \setminus \pm 1 \setminus \sqrt{143}}{- 2}$

$\setminus \implies x \setminus \approx - 11.47913 \setminus q \quad , \setminus q \quad x \setminus \approx 0.47913$