(4x-4)(3x+17)=180

Solution for (4x-4)(3x+17)=180 equation:

(4x-4)(3x+17)=180

We move all terms to the left:

(4x-4)(3x+17)-(180)=0

We multiply parentheses ..

(+12x^2+68x-12x-68)-180=0

We get rid of parentheses

12x^2+68x-12x-68-180=0

We add all the numbers together, and all the variables

12x^2+56x-248=0

a = 12; b = 56; c = -248;
Δ = b2-4ac
Δ = 562-4·12·(-248)
Δ = 15040
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:

$\sqrt{\Delta}=\sqrt{15040}=\sqrt{64*235}=\sqrt{64}*\sqrt{235}=8\sqrt{235}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(56)-8\sqrt{235}}{2*12}=\frac{-56-8\sqrt{235}}{24}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(56)+8\sqrt{235}}{2*12}=\frac{-56+8\sqrt{235}}{24}
$