8=x^2+x-2+x^2+3x-12

Solution for 8=x^2+x-2+x^2+3x-12 equation:

8=x^2+x-2+x^2+3x-12

We move all terms to the left:

8-(x^2+x-2+x^2+3x-12)=0

We get rid of parentheses

-x^2-x^2-x-3x+2+12+8=0

We add all the numbers together, and all the variables

-2x^2-4x+22=0

a = -2; b = -4; c = +22;
Δ = b2-4ac
Δ = -42-4·(-2)·22
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{-(-4)-8\sqrt{3}}{2*-2}=\frac{4-8\sqrt{3}}{-4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+8\sqrt{3}}{2*-2}=\frac{4+8\sqrt{3}}{-4}
$