(x^2+x^2)+4(x^2+x)-12=0

Solution for (x^2+x^2)+4(x^2+x)-12=0 equation:

(x^2+x^2)+4(x^2+x)-12=0

We multiply parentheses

(x^2+x^2)+4x^2+4x-12=0

We get rid of parentheses

x^2+x^2+4x^2+4x-12=0

We add all the numbers together, and all the variables

6x^2+4x-12=0

a = 6; b = 4; c = -12;
Δ = b2-4ac
Δ = 42-4·6·(-12)
Δ = 304
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:

$\sqrt{\Delta}=\sqrt{304}=\sqrt{16*19}=\sqrt{16}*\sqrt{19}=4\sqrt{19}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{19}}{2*6}=\frac{-4-4\sqrt{19}}{12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{19}}{2*6}=\frac{-4+4\sqrt{19}}{12}
$