2(x^2)+4(x^2+3)-2=22

Solution for 2(x^2)+4(x^2+3)-2=22 equation:

2(x^2)+4(x^2+3)-2=22

We move all terms to the left:

2(x^2)+4(x^2+3)-2-(22)=0

We add all the numbers together, and all the variables

2x^2+4(x^2+3)-24=0

We multiply parentheses

2x^2+4x^2+12-24=0

We add all the numbers together, and all the variables

6x^2-12=0

a = 6; b = 0; c = -12;
Δ = b2-4ac
Δ = 02-4·6·(-12)
Δ = 288
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:

$\sqrt{\Delta}=\sqrt{288}=\sqrt{144*2}=\sqrt{144}*\sqrt{2}=12\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-12\sqrt{2}}{2*6}=\frac{0-12\sqrt{2}}{12}
=-\frac{12\sqrt{2}}{12}
=-\sqrt{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+12\sqrt{2}}{2*6}=\frac{0+12\sqrt{2}}{12}
=\frac{12\sqrt{2}}{12}
=\sqrt{2}
$