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

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

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

We move all terms to the left:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

2x^2+4x-16=0

a = 2; b = 4; c = -16;
Δ = b2-4ac
Δ = 42-4·2·(-16)
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-12}{2*2}=\frac{-16}{4}
=-4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+12}{2*2}=\frac{8}{4}
=2
$