100+y^2=3y^2

Solution for 100+y^2=3y^2 equation:

100+y^2=3y^2

We move all terms to the left:

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

We add all the numbers together, and all the variables

-2y^2+100=0

a = -2; b = 0; c = +100;
Δ = b2-4ac
Δ = 02-4·(-2)·100
Δ = 800
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{800}=\sqrt{400*2}=\sqrt{400}*\sqrt{2}=20\sqrt{2}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-20\sqrt{2}}{2*-2}=\frac{0-20\sqrt{2}}{-4}
=-\frac{20\sqrt{2}}{-4}
=-\frac{5\sqrt{2}}{-1}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+20\sqrt{2}}{2*-2}=\frac{0+20\sqrt{2}}{-4}
=\frac{20\sqrt{2}}{-4}
=\frac{5\sqrt{2}}{-1}
$