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

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

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

We move all terms to the left:

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


We calculate terms in parentheses: -(3(x^2-6)+2), so:

3(x^2-6)+2

We multiply parentheses

3x^2-18+2

We add all the numbers together, and all the variables

3x^2-16

Back to the equation:

-(3x^2-16)


We get rid of parentheses

2x^2-3x^2+16=0

We add all the numbers together, and all the variables

-1x^2+16=0

a = -1; b = 0; c = +16;
Δ = b2-4ac
Δ = 02-4·(-1)·16
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*-1}=\frac{-8}{-2}
=+4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8}{2*-1}=\frac{8}{-2}
=-4
$