X^2+(3x^2)=64

Solution for X^2+(3x^2)=64 equation:

X^2+(3X^2)=64

We move all terms to the left:

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

We add all the numbers together, and all the variables

4X^2-64=0

a = 4; b = 0; c = -64;
Δ = b2-4ac
Δ = 02-4·4·(-64)
Δ = 1024
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1024}=32$
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-32}{2*4}=\frac{-32}{8}
=-4
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+32}{2*4}=\frac{32}{8}
=4
$