3(2-x)(x+2)=x2

Solution for 3(2-x)(x+2)=x2 equation:

3(2-x)(x+2)=x2

We move all terms to the left:

3(2-x)(x+2)-(x2)=0

We add all the numbers together, and all the variables

3(-1x+2)(x+2)-x2=0

We add all the numbers together, and all the variables

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

We multiply parentheses ..

-1x^2+3(-1x^2-2x+2x+4)=0

We multiply parentheses

-1x^2-3x^2-6x+6x+12=0

We add all the numbers together, and all the variables

-4x^2+12=0

a = -4; b = 0; c = +12;
Δ = b2-4ac
Δ = 02-4·(-4)·12
Δ = 192
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:

$\sqrt{\Delta}=\sqrt{192}=\sqrt{64*3}=\sqrt{64}*\sqrt{3}=8\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{3}}{2*-4}=\frac{0-8\sqrt{3}}{-8}
=-\frac{8\sqrt{3}}{-8}
=-\frac{\sqrt{3}}{-1}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{3}}{2*-4}=\frac{0+8\sqrt{3}}{-8}
=\frac{8\sqrt{3}}{-8}
=\frac{\sqrt{3}}{-1}
$