3x^2+12x-152=0

Solution for 3x^2+12x-152=0 equation:

3x^2+12x-152=0

a = 3; b = 12; c = -152;
Δ = b2-4ac
Δ = 122-4·3·(-152)
Δ = 1968
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:

$\sqrt{\Delta}=\sqrt{1968}=\sqrt{16*123}=\sqrt{16}*\sqrt{123}=4\sqrt{123}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-4\sqrt{123}}{2*3}=\frac{-12-4\sqrt{123}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+4\sqrt{123}}{2*3}=\frac{-12+4\sqrt{123}}{6}
$