2x^2+12=4x^2+4

Solution for 2x^2+12=4x^2+4 equation:

2x^2+12=4x^2+4

We move all terms to the left:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

-2x^2+8=0

a = -2; b = 0; c = +8;
Δ = b2-4ac
Δ = 02-4·(-2)·8
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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