12x^2-4x-96=0

Solution for 12x^2-4x-96=0 equation:

12x^2-4x-96=0

a = 12; b = -4; c = -96;
Δ = b2-4ac
Δ = -42-4·12·(-96)
Δ = 4624
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4624}=68$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-68}{2*12}=\frac{-64}{24}
=-2+2/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+68}{2*12}=\frac{72}{24}
=3
$