3x^2+24x-156=0

Solution for 3x^2+24x-156=0 equation:

3x^2+24x-156=0

a = 3; b = 24; c = -156;
Δ = b2-4ac
Δ = 242-4·3·(-156)
Δ = 2448
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:

$\sqrt{\Delta}=\sqrt{2448}=\sqrt{144*17}=\sqrt{144}*\sqrt{17}=12\sqrt{17}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-12\sqrt{17}}{2*3}=\frac{-24-12\sqrt{17}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+12\sqrt{17}}{2*3}=\frac{-24+12\sqrt{17}}{6}
$