360=(4x-4)(3x+2)

Solution for 360=(4x-4)(3x+2) equation:

360=(4x-4)(3x+2)

We move all terms to the left:

360-((4x-4)(3x+2))=0

We multiply parentheses ..

-((+12x^2+8x-12x-8))+360=0


We calculate terms in parentheses: -((+12x^2+8x-12x-8)), so:

(+12x^2+8x-12x-8)

We get rid of parentheses

12x^2+8x-12x-8

We add all the numbers together, and all the variables

12x^2-4x-8

Back to the equation:

-(12x^2-4x-8)


We get rid of parentheses

-12x^2+4x+8+360=0

We add all the numbers together, and all the variables

-12x^2+4x+368=0

a = -12; b = 4; c = +368;
Δ = b2-4ac
Δ = 42-4·(-12)·368
Δ = 17680
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:

$\sqrt{\Delta}=\sqrt{17680}=\sqrt{16*1105}=\sqrt{16}*\sqrt{1105}=4\sqrt{1105}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{1105}}{2*-12}=\frac{-4-4\sqrt{1105}}{-24}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{1105}}{2*-12}=\frac{-4+4\sqrt{1105}}{-24}
$