# How do you find the roots, real and imaginary, of y=(x/5-1)(-2x+5) using the quadratic formula?

Jan 15, 2017

$x \in \left\{5 , \frac{5}{2}\right\}$

#### Explanation:

No need to use the quadratic formula to find the roots since the equation is already factored out. Equate y to 0 to find the roots.

$y = 0 = \left(\frac{x}{5} - 1\right) \left(- 2 x + 5\right)$

$\implies \frac{x}{5} - 1 = 0$, and/or $- 2 x + 5 = 0$

$\implies \frac{x}{5} = 1 \implies x = 5$

$\implies - 2 x = - 5 \implies x = \frac{5}{2}$

If you really need to use the quadratic formula

$y = 0 = \left(\frac{x}{5} - 1\right) \left(- 2 x + 5\right)$

$\implies 0 = - \frac{2}{5} {x}^{2} + 3 x - 5$

-

$a {x}^{2} + b x + c = 0$, its roots can be determined by the formula

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

=> x = (-3 +- sqrt(3^2 - 4*(-2/5)(-5)))/(2(-2/5)

Proceeding with the computation should give us the same result as with the one above.