3x(X+1)=96

Solution for 3x(X+1)=96 equation:

3x(x+1)=96

We move all terms to the left:

3x(x+1)-(96)=0

We multiply parentheses

3x^2+3x-96=0

a = 3; b = 3; c = -96;
Δ = b2-4ac
Δ = 32-4·3·(-96)
Δ = 1161
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1161}=\sqrt{9*129}=\sqrt{9}*\sqrt{129}=3\sqrt{129}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-3\sqrt{129}}{2*3}=\frac{-3-3\sqrt{129}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+3\sqrt{129}}{2*3}=\frac{-3+3\sqrt{129}}{6}
$