# How do you find the roots, real and imaginary, of #y= 2x^2-x-(x+3)^2 # using the quadratic formula?

##### 1 Answer

#### Answer:

#### Explanation:

By roots, I will assume you mean zeros.

First expand out the quadratic, then combine terms to bring it into standard form:

#2x^2-x-(x+3)^2 = 2x^2-x-(x^2+6x+9)#

#color(white)(2x^2-x-(x+3)^2) = 2x^2-x-x^2-6x-9#

#color(white)(2x^2-x-(x+3)^2) = x^2-7x-9#

This is in the form:

#ax^2+bx+c#

with

It has discriminant

#Delta = b^2-4ac = (color(blue)(-7))^2-4(color(blue)(1))(color(blue)(-9)) = 49+36 = 85#

Since

We can use the quadratic formula to find them:

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

#color(white)(x) = (-b+-sqrt(Delta))/(2a)#

#color(white)(x) = (7+-sqrt(85))/2#

#color(white)(x) = 7/2+-sqrt(85)/2#