0=y^2-13y+32

Solution for 0=y^2-13y+32 equation:

0=y^2-13y+32

We move all terms to the left:

0-(y^2-13y+32)=0

We add all the numbers together, and all the variables

-(y^2-13y+32)=0

We get rid of parentheses

-y^2+13y-32=0

We add all the numbers together, and all the variables

-1y^2+13y-32=0

a = -1; b = 13; c = -32;
Δ = b2-4ac
Δ = 132-4·(-1)·(-32)
Δ = 41
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{-(13)-\sqrt{41}}{2*-1}=\frac{-13-\sqrt{41}}{-2}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+\sqrt{41}}{2*-1}=\frac{-13+\sqrt{41}}{-2}
$