2(3x^2+4x-48)=0

Solution for 2(3x^2+4x-48)=0 equation:

2(3x^2+4x-48)=0

We multiply parentheses

6x^2+8x-96=0

a = 6; b = 8; c = -96;
Δ = b2-4ac
Δ = 82-4·6·(-96)
Δ = 2368
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:

$\sqrt{\Delta}=\sqrt{2368}=\sqrt{64*37}=\sqrt{64}*\sqrt{37}=8\sqrt{37}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-8\sqrt{37}}{2*6}=\frac{-8-8\sqrt{37}}{12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+8\sqrt{37}}{2*6}=\frac{-8+8\sqrt{37}}{12}
$