3x^2+120x-800=0

Solution for 3x^2+120x-800=0 equation:

3x^2+120x-800=0

a = 3; b = 120; c = -800;
Δ = b2-4ac
Δ = 1202-4·3·(-800)
Δ = 24000
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:

$\sqrt{\Delta}=\sqrt{24000}=\sqrt{1600*15}=\sqrt{1600}*\sqrt{15}=40\sqrt{15}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(120)-40\sqrt{15}}{2*3}=\frac{-120-40\sqrt{15}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(120)+40\sqrt{15}}{2*3}=\frac{-120+40\sqrt{15}}{6}
$