(2x)(3x+20)=85+65

Solution for (2x)(3x+20)=85+65 equation:

(2x)(3x+20)=85+65

We move all terms to the left:

(2x)(3x+20)-(85+65)=0

We add all the numbers together, and all the variables

2x(3x+20)-150=0

We multiply parentheses

6x^2+40x-150=0

a = 6; b = 40; c = -150;
Δ = b2-4ac
Δ = 402-4·6·(-150)
Δ = 5200
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:

$\sqrt{\Delta}=\sqrt{5200}=\sqrt{400*13}=\sqrt{400}*\sqrt{13}=20\sqrt{13}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-20\sqrt{13}}{2*6}=\frac{-40-20\sqrt{13}}{12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+20\sqrt{13}}{2*6}=\frac{-40+20\sqrt{13}}{12}
$