-13y^2+36=-y^2

Solution for -13y^2+36=-y^2 equation:

-13y^2+36=-y^2

We move all terms to the left:

-13y^2+36-(-y^2)=0

We get rid of parentheses

-13y^2+y^2+36=0

We add all the numbers together, and all the variables

-12y^2+36=0

a = -12; b = 0; c = +36;
Δ = b2-4ac
Δ = 02-4·(-12)·36
Δ = 1728
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:

$\sqrt{\Delta}=\sqrt{1728}=\sqrt{576*3}=\sqrt{576}*\sqrt{3}=24\sqrt{3}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-24\sqrt{3}}{2*-12}=\frac{0-24\sqrt{3}}{-24}
=-\frac{24\sqrt{3}}{-24}
=-\frac{\sqrt{3}}{-1}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+24\sqrt{3}}{2*-12}=\frac{0+24\sqrt{3}}{-24}
=\frac{24\sqrt{3}}{-24}
=\frac{\sqrt{3}}{-1}
$