(3x-38)(2x+12)=180

Solution for (3x-38)(2x+12)=180 equation:

(3x-38)(2x+12)=180

We move all terms to the left:

(3x-38)(2x+12)-(180)=0

We multiply parentheses ..

(+6x^2+36x-76x-456)-180=0

We get rid of parentheses

6x^2+36x-76x-456-180=0

We add all the numbers together, and all the variables

6x^2-40x-636=0

a = 6; b = -40; c = -636;
Δ = b2-4ac
Δ = -402-4·6·(-636)
Δ = 16864
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:

$\sqrt{\Delta}=\sqrt{16864}=\sqrt{16*1054}=\sqrt{16}*\sqrt{1054}=4\sqrt{1054}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-4\sqrt{1054}}{2*6}=\frac{40-4\sqrt{1054}}{12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+4\sqrt{1054}}{2*6}=\frac{40+4\sqrt{1054}}{12}
$