3y^2=16y

Solution for 3y^2=16y equation:

3y^2=16y

We move all terms to the left:

3y^2-(16y)=0

a = 3; b = -16; c = 0;
Δ = b2-4ac
Δ = -162-4·3·0
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*3}=\frac{0}{6}
=0
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+16}{2*3}=\frac{32}{6}
=5+1/3
$