3x2-96x+24=0

Solution for 3x2-96x+24=0 equation:

3x^2-96x+24=0

a = 3; b = -96; c = +24;
Δ = b2-4ac
Δ = -962-4·3·24
Δ = 8928
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:

$\sqrt{\Delta}=\sqrt{8928}=\sqrt{144*62}=\sqrt{144}*\sqrt{62}=12\sqrt{62}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-96)-12\sqrt{62}}{2*3}=\frac{96-12\sqrt{62}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-96)+12\sqrt{62}}{2*3}=\frac{96+12\sqrt{62}}{6}
$