4x(x+3)=96

Solution for 4x(x+3)=96 equation:

4x(x+3)=96

We move all terms to the left:

4x(x+3)-(96)=0

We multiply parentheses

4x^2+12x-96=0

a = 4; b = 12; c = -96;
Δ = b2-4ac
Δ = 122-4·4·(-96)
Δ = 1680
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:

$\sqrt{\Delta}=\sqrt{1680}=\sqrt{16*105}=\sqrt{16}*\sqrt{105}=4\sqrt{105}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-4\sqrt{105}}{2*4}=\frac{-12-4\sqrt{105}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+4\sqrt{105}}{2*4}=\frac{-12+4\sqrt{105}}{8}
$