2x(x-3)+(x-6)(2x+8)=0

Solution for 2x(x-3)+(x-6)(2x+8)=0 equation:

2x(x-3)+(x-6)(2x+8)=0

We multiply parentheses

2x^2-6x+(x-6)(2x+8)=0

We multiply parentheses ..

2x^2+(+2x^2+8x-12x-48)-6x=0

We get rid of parentheses

2x^2+2x^2+8x-12x-6x-48=0

We add all the numbers together, and all the variables

4x^2-10x-48=0

a = 4; b = -10; c = -48;
Δ = b2-4ac
Δ = -102-4·4·(-48)
Δ = 868
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:

$\sqrt{\Delta}=\sqrt{868}=\sqrt{4*217}=\sqrt{4}*\sqrt{217}=2\sqrt{217}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-2\sqrt{217}}{2*4}=\frac{10-2\sqrt{217}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+2\sqrt{217}}{2*4}=\frac{10+2\sqrt{217}}{8}
$