(16x^2)+(4x^2)=180

Solution for (16x^2)+(4x^2)=180 equation:

(16x^2)+(4x^2)=180

We move all terms to the left:

(16x^2)+(4x^2)-(180)=0

We add all the numbers together, and all the variables

20x^2-180=0

a = 20; b = 0; c = -180;
Δ = b2-4ac
Δ = 02-4·20·(-180)
Δ = 14400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{14400}=120$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-120}{2*20}=\frac{-120}{40}
=-3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+120}{2*20}=\frac{120}{40}
=3
$