# How do you find the roots, real and imaginary, of #y=(-2x+4)(x+1)+3x^2+2x # using the quadratic formula?

##### 1 Answer

#### Answer:

The root is

#### Explanation:

Convert the right-hand side to the form

#y=" "(–2x+4)(x+1)" " + 3x^2 + 2x#

#color(white)y=–2x^2-2x+4x+4 + 3x^2 + 2x#

#color(white)y=x^2+4x+4#

The roots of this function occur when

#0=x^2+4x+4#

Using the quadratic formula, the solutions for a quadratic

#x=(–b+-sqrt(b^2-4ac))/(2a)#

For the given quadratic,

#x=(–4+-sqrt(4^2-4(1)(4)))/(2(1))#

#x=(–4+-sqrt(16-16))/2#

#x=(–4+-sqrt(0))/2#

#x=(–4)/2" " = –2#

But wait—shouldn't we get **two** roots?

Yes, we should, and we did! Since the discriminant **multiplicity** of two. Meaning, the factor

Indeed, we can confirm this if we factor

#x^2+4x+4 = (x+2)(x+2).#