# How do you find the roots, real and imaginary, of y= (45-x)x- (x-2)(2x-1)  using the quadratic formula?

Feb 24, 2018

$x = \frac{25}{3} \pm \frac{\sqrt{619}}{3}$

#### Explanation:

Note that the quadratic equation is given in the form of a function, so the question should really be asking about zeros, not roots.

Proceeding on the assumption that you want to solve for $y = 0$...

First we want to get this quadratic into standard form, which we can do by multiplying out and combining terms:

$\left(45 - x\right) x - \left(x - 2\right) \left(2 x - 1\right) = \left(45 x - {x}^{2}\right) - \left(2 {x}^{2} - 5 x + 2\right)$

$\textcolor{w h i t e}{\left(45 - x\right) x - \left(x - 2\right) \left(2 x - 1\right)} = - 3 {x}^{2} + 50 x - 2$

This is now in the form:

$a {x}^{2} + b x + c$

with $a = - 3$, $b = 50$ and $c = - 2$

It has zeros given by the quadratic formula:

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

color(white)(x) = (-(color(blue)(50))+-sqrt((color(blue)(50))^2-4(color(blue)(-3))(color(blue)(-2))))/(2(color(blue)(-3))

$\textcolor{w h i t e}{x} = \frac{- 50 \pm \sqrt{2500 - 24}}{- 6}$

$\textcolor{w h i t e}{x} = \frac{50 \pm \sqrt{2476}}{6}$

$\textcolor{w h i t e}{x} = \frac{50 \pm \sqrt{{2}^{2} \cdot 619}}{6}$

$\textcolor{w h i t e}{x} = \frac{50 \pm 2 \sqrt{619}}{6}$

$\textcolor{w h i t e}{x} = \frac{25}{3} \pm \frac{\sqrt{619}}{3}$