3x^2-62x-240=0

Solution for 3x^2-62x-240=0 equation:

3x^2-62x-240=0

a = 3; b = -62; c = -240;
Δ = b2-4ac
Δ = -622-4·3·(-240)
Δ = 6724
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{6724}=82$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-62)-82}{2*3}=\frac{-20}{6}
=-3+1/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-62)+82}{2*3}=\frac{144}{6}
=24
$