2(2x^2-5)+12=3x^2+6

Solution for 2(2x^2-5)+12=3x^2+6 equation:

2(2x^2-5)+12=3x^2+6

We move all terms to the left:

2(2x^2-5)+12-(3x^2+6)=0

We multiply parentheses

4x^2-(3x^2+6)-10+12=0

We get rid of parentheses

4x^2-3x^2-6-10+12=0

We add all the numbers together, and all the variables

x^2-4=0

a = 1; b = 0; c = -4;
Δ = b2-4ac
Δ = 02-4·1·(-4)
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4}{2*1}=\frac{-4}{2}
=-2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4}{2*1}=\frac{4}{2}
=2
$