3x^2+36x-60=0

Solution for 3x^2+36x-60=0 equation:

3x^2+36x-60=0

a = 3; b = 36; c = -60;
Δ = b2-4ac
Δ = 362-4·3·(-60)
Δ = 2016
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:

$\sqrt{\Delta}=\sqrt{2016}=\sqrt{144*14}=\sqrt{144}*\sqrt{14}=12\sqrt{14}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(36)-12\sqrt{14}}{2*3}=\frac{-36-12\sqrt{14}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(36)+12\sqrt{14}}{2*3}=\frac{-36+12\sqrt{14}}{6}
$