A=(2x+2)(3x)

Solution for A=(2x+2)(3x) equation:

=(2A+2)(3A)

We move all terms to the left:

-((2A+2)(3A))=0


We calculate terms in parentheses: -((2A+2)3A), so:

(2A+2)3A

We multiply parentheses

6A^2+6A

Back to the equation:

-(6A^2+6A)


We get rid of parentheses

-6A^2-6A=0

a = -6; b = -6; c = 0;
Δ = b2-4ac
Δ = -62-4·(-6)·0
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$A_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$A_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$A_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-6}{2*-6}=\frac{0}{-12}
=0
$
$A_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+6}{2*-6}=\frac{12}{-12}
=-1
$