3y*(2+y)=336

Solution for 3y*(2+y)=336 equation:

3y(2+y)=336

We move all terms to the left:

3y(2+y)-(336)=0

We add all the numbers together, and all the variables

3y(y+2)-336=0

We multiply parentheses

3y^2+6y-336=0

a = 3; b = 6; c = -336;
Δ = b2-4ac
Δ = 62-4·3·(-336)
Δ = 4068
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{4068}=\sqrt{36*113}=\sqrt{36}*\sqrt{113}=6\sqrt{113}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-6\sqrt{113}}{2*3}=\frac{-6-6\sqrt{113}}{6}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+6\sqrt{113}}{2*3}=\frac{-6+6\sqrt{113}}{6}
$