0=-y^2+16y

Solution for 0=-y^2+16y equation:

0=-y^2+16y

We move all terms to the left:

0-(-y^2+16y)=0

We add all the numbers together, and all the variables

-(-y^2+16y)=0

We get rid of parentheses

y^2-16y=0

a = 1; b = -16; c = 0;
Δ = b2-4ac
Δ = -162-4·1·0
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-16}{2*1}=\frac{0}{2}
=0
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+16}{2*1}=\frac{32}{2}
=16
$