(2x+2)(3x+3)=2x+50

Solution for (2x+2)(3x+3)=2x+50 equation:

(2x+2)(3x+3)=2x+50

We move all terms to the left:

(2x+2)(3x+3)-(2x+50)=0

We get rid of parentheses

(2x+2)(3x+3)-2x-50=0

We multiply parentheses ..

(+6x^2+6x+6x+6)-2x-50=0

We get rid of parentheses

6x^2+6x+6x-2x+6-50=0

We add all the numbers together, and all the variables

6x^2+10x-44=0

a = 6; b = 10; c = -44;
Δ = b2-4ac
Δ = 102-4·6·(-44)
Δ = 1156
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1156}=34$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-34}{2*6}=\frac{-44}{12}
=-3+2/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+34}{2*6}=\frac{24}{12}
=2
$