0=y^2+15y-36

Solution for 0=y^2+15y-36 equation:

0=y^2+15y-36

We move all terms to the left:

0-(y^2+15y-36)=0

We add all the numbers together, and all the variables

-(y^2+15y-36)=0

We get rid of parentheses

-y^2-15y+36=0

We add all the numbers together, and all the variables

-1y^2-15y+36=0

a = -1; b = -15; c = +36;
Δ = b2-4ac
Δ = -152-4·(-1)·36
Δ = 369
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:

$\sqrt{\Delta}=\sqrt{369}=\sqrt{9*41}=\sqrt{9}*\sqrt{41}=3\sqrt{41}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-15)-3\sqrt{41}}{2*-1}=\frac{15-3\sqrt{41}}{-2}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+3\sqrt{41}}{2*-1}=\frac{15+3\sqrt{41}}{-2}
$