3y2=8y+16

Solution for 3y2=8y+16 equation:

3y^2=8y+16

We move all terms to the left:

3y^2-(8y+16)=0

We get rid of parentheses

3y^2-8y-16=0

a = 3; b = -8; c = -16;
Δ = b2-4ac
Δ = -82-4·3·(-16)
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{-(-8)-16}{2*3}=\frac{-8}{6}
=-1+1/3
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+16}{2*3}=\frac{24}{6}
=4
$