3x^2+36x-4860=0

Solution for 3x^2+36x-4860=0 equation:

3x^2+36x-4860=0

a = 3; b = 36; c = -4860;
Δ = b2-4ac
Δ = 362-4·3·(-4860)
Δ = 59616
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:

$\sqrt{\Delta}=\sqrt{59616}=\sqrt{1296*46}=\sqrt{1296}*\sqrt{46}=36\sqrt{46}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(36)-36\sqrt{46}}{2*3}=\frac{-36-36\sqrt{46}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(36)+36\sqrt{46}}{2*3}=\frac{-36+36\sqrt{46}}{6}
$