(2y+3)(3y)=96

Solution for (2y+3)(3y)=96 equation:

(2y+3)(3y)=96

We move all terms to the left:

(2y+3)(3y)-(96)=0

We multiply parentheses

6y^2+9y-96=0

a = 6; b = 9; c = -96;
Δ = b2-4ac
Δ = 92-4·6·(-96)
Δ = 2385
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:

$\sqrt{\Delta}=\sqrt{2385}=\sqrt{9*265}=\sqrt{9}*\sqrt{265}=3\sqrt{265}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-3\sqrt{265}}{2*6}=\frac{-9-3\sqrt{265}}{12}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+3\sqrt{265}}{2*6}=\frac{-9+3\sqrt{265}}{12}
$