# How do you find the roots, real and imaginary, of y=(x/5-5)(-3x-2) using the quadratic formula?

How do you find the roots, real and imaginary, of y=(x/5-5)(-3x-2) using the quadratic formula?

ReportAnswer 1

See a solution process below:

#### Explanation:

First, we need to write this equation in standard form:

$y = \left(\textcolor{red}{\frac{x}{5}} - \textcolor{red}{5}\right) \left(\textcolor{b l u e}{- 3 x} - \textcolor{b l u e}{2}\right)$ becomes:

$y = - \left(\textcolor{red}{\frac{x}{5}} \times \textcolor{b l u e}{3 x}\right) - \left(\textcolor{red}{\frac{x}{5}} \times \textcolor{b l u e}{2}\right) + \left(\textcolor{red}{5} \times \textcolor{b l u e}{3 x}\right) + \left(\textcolor{red}{5} \times \textcolor{b l u e}{2}\right)$

$y = - \frac{3}{5} {x}^{2} - \frac{2}{5} x + 15 x + 10$

$y = - \frac{3}{5} {x}^{2} - \frac{2}{5} x + \left(\frac{5}{5} \cdot 15\right) x + 10$

$y = - \frac{3}{5} {x}^{2} - \frac{2}{5} x + \frac{75}{5} x + 10$

$y = - \frac{3}{5} {x}^{2} + \frac{73}{5} x + 10$

We can now use the quadratic equation to solve the equation. The quadratic formula states:

For $\textcolor{red}{a} {x}^{2} + \textcolor{b l u e}{b} x + \textcolor{g r e e n}{c} = 0$, the values of $x$ which are the solutions to the equation are given by:

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

Substituting:

$\textcolor{red}{- \frac{3}{5}}$ for $\textcolor{red}{a}$

$\textcolor{b l u e}{\frac{73}{5}}$ for $\textcolor{b l u e}{b}$

$\textcolor{g r e e n}{10}$ for $\textcolor{g r e e n}{c}$ gives:

$x = \frac{- \textcolor{b l u e}{\frac{73}{5}} \pm \sqrt{{\textcolor{b l u e}{\frac{73}{5}}}^{2} - \left(4 \cdot \textcolor{red}{- \frac{3}{5}} \cdot \textcolor{g r e e n}{10}\right)}}{2 \cdot \textcolor{red}{- \frac{3}{5}}}$

$x = \frac{- \textcolor{b l u e}{\frac{73}{5}} \pm \sqrt{\frac{5329}{25} - \left(- \frac{120}{5}\right)}}{- \frac{6}{5}}$

$x = \frac{- \textcolor{b l u e}{\frac{73}{5}} \pm \sqrt{\frac{5329}{25} - \left(\frac{5}{5} \cdot - \frac{120}{5}\right)}}{- \frac{6}{5}}$

$x = \frac{- \textcolor{b l u e}{\frac{73}{5}} \pm \sqrt{\frac{5329}{25} - \left(- \frac{600}{25}\right)}}{- \frac{6}{5}}$

$x = \frac{- \textcolor{b l u e}{\frac{73}{5}} \pm \sqrt{\frac{5329}{25} + \frac{600}{25}}}{- \frac{6}{5}}$

$x = \frac{- \textcolor{b l u e}{\frac{73}{5}} \pm \sqrt{\frac{5929}{25}}}{- \frac{6}{5}}$

$x = \frac{- \textcolor{b l u e}{\frac{73}{5}} + \frac{77}{5}}{- \frac{6}{5}}$ and $x = \frac{- \textcolor{b l u e}{\frac{73}{5}} - \frac{77}{5}}{- \frac{6}{5}}$

$x = \frac{\frac{4}{5}}{- \frac{6}{5}}$ and $x = \frac{- \frac{150}{5}}{- \frac{6}{5}}$

$x = - \frac{4 \cdot 5}{5 \cdot 6}$ and $x = \frac{150 \cdot 5}{5 \cdot 6}$

$x = - \frac{4}{6}$ and $x = \frac{150}{6}$

$x = - \frac{2}{3}$ and $x = 25$

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