(3x^2-3)=36

Solution for (3x^2-3)=36 equation:

(3x^2-3)=36

We move all terms to the left:

(3x^2-3)-(36)=0

We get rid of parentheses

3x^2-3-36=0

We add all the numbers together, and all the variables

3x^2-39=0

a = 3; b = 0; c = -39;
Δ = b2-4ac
Δ = 02-4·3·(-39)
Δ = 468
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:

$\sqrt{\Delta}=\sqrt{468}=\sqrt{36*13}=\sqrt{36}*\sqrt{13}=6\sqrt{13}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{13}}{2*3}=\frac{0-6\sqrt{13}}{6}
=-\frac{6\sqrt{13}}{6}
=-\sqrt{13}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{13}}{2*3}=\frac{0+6\sqrt{13}}{6}
=\frac{6\sqrt{13}}{6}
=\sqrt{13}
$