3x^2+12x-144=0

Solution for 3x^2+12x-144=0 equation:

3x^2+12x-144=0

a = 3; b = 12; c = -144;
Δ = b2-4ac
Δ = 122-4·3·(-144)
Δ = 1872
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:

$\sqrt{\Delta}=\sqrt{1872}=\sqrt{144*13}=\sqrt{144}*\sqrt{13}=12\sqrt{13}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-12\sqrt{13}}{2*3}=\frac{-12-12\sqrt{13}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+12\sqrt{13}}{2*3}=\frac{-12+12\sqrt{13}}{6}
$