3x^2=8x+96

Solution for 3x^2=8x+96 equation:

3x^2=8x+96

We move all terms to the left:

3x^2-(8x+96)=0

We get rid of parentheses

3x^2-8x-96=0

a = 3; b = -8; c = -96;
Δ = b2-4ac
Δ = -82-4·3·(-96)
Δ = 1216
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:

$\sqrt{\Delta}=\sqrt{1216}=\sqrt{64*19}=\sqrt{64}*\sqrt{19}=8\sqrt{19}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-8\sqrt{19}}{2*3}=\frac{8-8\sqrt{19}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+8\sqrt{19}}{2*3}=\frac{8+8\sqrt{19}}{6}
$