2y^2=+16y

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

2y^2=+16y

We move all terms to the left:

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

We get rid of parentheses

2y^2-16y=0

a = 2; b = -16; c = 0;
Δ = b2-4ac
Δ = -162-4·2·0
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*2}=\frac{0}{4}
=0
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+16}{2*2}=\frac{32}{4}
=8
$