# How do you find the roots, real and imaginary, of y= 5x^2 - 245  using the quadratic formula?

Oct 26, 2017

See a solution process below:

#### Explanation:

First, we can rewrite this equation as:

$y = 5 {x}^{2} + 0 x - 245$

To find the roots we need to set the right side of the equation equal to $0$ and solve for $x$:

$5 {x}^{2} + 0 x - 245 = 0$

We can now use the quadratic equation to solve this problem:

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:

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

Substituting:

$\textcolor{red}{5}$ for $\textcolor{red}{a}$

$\textcolor{b l u e}{0}$ for $\textcolor{b l u e}{b}$

$\textcolor{g r e e n}{- 245}$ for $\textcolor{g r e e n}{c}$ gives:

$x = \frac{- \textcolor{b l u e}{0} \pm \sqrt{{\textcolor{b l u e}{0}}^{2} - \left(4 \cdot \textcolor{red}{5} \cdot \textcolor{g r e e n}{- 245}\right)}}{2 \cdot \textcolor{red}{5}}$

$x = \pm \frac{\sqrt{0 - \left(- 4900\right)}}{10}$

$x = \pm \frac{\sqrt{0 + 4900}}{10}$

$x = \pm \frac{\sqrt{4900}}{10}$

$x = - \frac{70}{10}$ and $x = \frac{70}{10}$

$x = - 7$ and $x = 7$