3x^2+42x-117=0

Solution for 3x^2+42x-117=0 equation:

3x^2+42x-117=0

a = 3; b = 42; c = -117;
Δ = b2-4ac
Δ = 422-4·3·(-117)
Δ = 3168
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:

$\sqrt{\Delta}=\sqrt{3168}=\sqrt{144*22}=\sqrt{144}*\sqrt{22}=12\sqrt{22}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(42)-12\sqrt{22}}{2*3}=\frac{-42-12\sqrt{22}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(42)+12\sqrt{22}}{2*3}=\frac{-42+12\sqrt{22}}{6}
$