0=-y^2+y+32

Solution for 0=-y^2+y+32 equation:

0=-y^2+y+32

We move all terms to the left:

0-(-y^2+y+32)=0

We add all the numbers together, and all the variables

-(-y^2+y+32)=0

We get rid of parentheses

y^2-y-32=0

We add all the numbers together, and all the variables

y^2-1y-32=0

a = 1; b = -1; c = -32;
Δ = b2-4ac
Δ = -12-4·1·(-32)
Δ = 129
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-\sqrt{129}}{2*1}=\frac{1-\sqrt{129}}{2}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{129}}{2*1}=\frac{1+\sqrt{129}}{2}
$