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

##### 1 Answer

#### Answer:

#### 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

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

#(45-x)x-(x-2)(2x-1) = (45x-x^2)-(2x^2-5x+2)#

#color(white)((45-x)x-(x-2)(2x-1)) = -3x^2+50x-2#

This is now in the form:

#ax^2+bx+c#

with

It has zeros given by the quadratic formula:

#x = (-b+-sqrt(b^2-4ac))/(2a)#

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

#color(white)(x) = (-50+-sqrt(2500-24))/(-6)#

#color(white)(x) = (50+-sqrt(2476))/6#

#color(white)(x) = (50+-sqrt(2^2 * 619))/6#

#color(white)(x) = (50+-2sqrt(619))/6#

#color(white)(x) = 25/3+-sqrt(619)/3#