3x^2=x^2+64

Solution for 3x^2=x^2+64 equation:

3x^2=x^2+64

We move all terms to the left:

3x^2-(x^2+64)=0

We get rid of parentheses

3x^2-x^2-64=0

We add all the numbers together, and all the variables

2x^2-64=0

a = 2; b = 0; c = -64;
Δ = b2-4ac
Δ = 02-4·2·(-64)
Δ = 512
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:

$\sqrt{\Delta}=\sqrt{512}=\sqrt{256*2}=\sqrt{256}*\sqrt{2}=16\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16\sqrt{2}}{2*2}=\frac{0-16\sqrt{2}}{4}
=-\frac{16\sqrt{2}}{4}
=-4\sqrt{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16\sqrt{2}}{2*2}=\frac{0+16\sqrt{2}}{4}
=\frac{16\sqrt{2}}{4}
=4\sqrt{2}
$