(x^2)+((3x)^2)=100

Solution for (x^2)+((3x)^2)=100 equation:

(x^2)+((3x)^2)=100

We move all terms to the left:

(x^2)+((3x)^2)-(100)=0

determiningTheFunctionDomain
x^2+3x^2-100=0

We add all the numbers together, and all the variables

4x^2-100=0

a = 4; b = 0; c = -100;
Δ = b2-4ac
Δ = 02-4·4·(-100)
Δ = 1600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1600}=40$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-40}{2*4}=\frac{-40}{8}
=-5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+40}{2*4}=\frac{40}{8}
=5
$