-2(x-3)(x+1)=6

Solution for -2(x-3)(x+1)=6 equation:

-2(x-3)(x+1)=6

We move all terms to the left:

-2(x-3)(x+1)-(6)=0

We multiply parentheses ..

-2(+x^2+x-3x-3)-6=0

We multiply parentheses

-2x^2-2x+6x+6-6=0

We add all the numbers together, and all the variables

-2x^2+4x=0

a = -2; b = 4; c = 0;
Δ = b2-4ac
Δ = 42-4·(-2)·0
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4}{2*-2}=\frac{-8}{-4}
=+2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4}{2*-2}=\frac{0}{-4}
=0
$