(2x+2)(3x)=80

Solution for (2x+2)(3x)=80 equation:

(2x+2)(3x)=80

We move all terms to the left:

(2x+2)(3x)-(80)=0

We multiply parentheses

6x^2+6x-80=0

a = 6; b = 6; c = -80;
Δ = b2-4ac
Δ = 62-4·6·(-80)
Δ = 1956
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1956}=\sqrt{4*489}=\sqrt{4}*\sqrt{489}=2\sqrt{489}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{489}}{2*6}=\frac{-6-2\sqrt{489}}{12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{489}}{2*6}=\frac{-6+2\sqrt{489}}{12}
$