3x^2-18x-140=0

Solution for 3x^2-18x-140=0 equation:

3x^2-18x-140=0

a = 3; b = -18; c = -140;
Δ = b2-4ac
Δ = -182-4·3·(-140)
Δ = 2004
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:

$\sqrt{\Delta}=\sqrt{2004}=\sqrt{4*501}=\sqrt{4}*\sqrt{501}=2\sqrt{501}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-2\sqrt{501}}{2*3}=\frac{18-2\sqrt{501}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+2\sqrt{501}}{2*3}=\frac{18+2\sqrt{501}}{6}
$