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

May 1, 2018

${x}_{1} = 25 \mathmr{and} {x}_{2} = - 4$ are the two real roots of the given function.

#### Explanation:

$= \left(\frac{x}{5}\right) \left(- \frac{x}{2} - 2\right) - 5 \left(- \frac{x}{2} - 2\right)$
$= - {x}^{2} / 10 - \frac{2 x}{5} + \frac{5 x}{2} + 10$
$= \frac{- {x}^{2} - 4 x + 25 x + 100}{10}$

Note that by definition, the roots equal 0. So we can set $y = 0$
$\implies 0 = \frac{- {x}^{2} - 4 x + 25 x + 100}{10}$
$\implies \left(- {x}^{2} - 4 x + 25 x + 100\right) = 0$
Let's multiply this by -1 to simply things.
$\left({x}^{2} + 4 x - 25 x - 100\right) = 0$
Funnily, you can quite easily factorize this polynomial: by factoring x and -25. However, you need to use the formula, which complicates things.
$\left({x}^{2} - 21 x - 100\right) = 0$ The formula is ${x}_{1} = \frac{- b + \sqrt{{b}^{2} - 4 a c}}{2 a}$
${x}_{2} = \frac{- b - \sqrt{{b}^{2} - 4 a c}}{2 a}$

Substitute $a = 1 , b = - 21 \mathmr{and} c = - 100$ into both of these to get the two roots. I will not solve it here as it would clutter the answer, however, the end result would be ${x}_{1} = 25 \mathmr{and} {x}_{2} = - 4$.