12x^2-32=4x^2+8

Solution for 12x^2-32=4x^2+8 equation:

12x^2-32=4x^2+8

We move all terms to the left:

12x^2-32-(4x^2+8)=0

We get rid of parentheses

12x^2-4x^2-8-32=0

We add all the numbers together, and all the variables

8x^2-40=0

a = 8; b = 0; c = -40;
Δ = b2-4ac
Δ = 02-4·8·(-40)
Δ = 1280
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:

$\sqrt{\Delta}=\sqrt{1280}=\sqrt{256*5}=\sqrt{256}*\sqrt{5}=16\sqrt{5}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16\sqrt{5}}{2*8}=\frac{0-16\sqrt{5}}{16}
=-\frac{16\sqrt{5}}{16}
=-\sqrt{5}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16\sqrt{5}}{2*8}=\frac{0+16\sqrt{5}}{16}
=\frac{16\sqrt{5}}{16}
=\sqrt{5}
$