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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

-x^2-2x^2+36+3^2=0

We add all the numbers together, and all the variables

-3x^2+45=0

a = -3; b = 0; c = +45;
Δ = b2-4ac
Δ = 02-4·(-3)·45
Δ = 540
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:

$\sqrt{\Delta}=\sqrt{540}=\sqrt{36*15}=\sqrt{36}*\sqrt{15}=6\sqrt{15}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{15}}{2*-3}=\frac{0-6\sqrt{15}}{-6}
=-\frac{6\sqrt{15}}{-6}
=-\frac{\sqrt{15}}{-1}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{15}}{2*-3}=\frac{0+6\sqrt{15}}{-6}
=\frac{6\sqrt{15}}{-6}
=\frac{\sqrt{15}}{-1}
$