3x^2+24x-40=0

Solution for 3x^2+24x-40=0 equation:

3x^2+24x-40=0

a = 3; b = 24; c = -40;
Δ = b2-4ac
Δ = 242-4·3·(-40)
Δ = 1056
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:

$\sqrt{\Delta}=\sqrt{1056}=\sqrt{16*66}=\sqrt{16}*\sqrt{66}=4\sqrt{66}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-4\sqrt{66}}{2*3}=\frac{-24-4\sqrt{66}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+4\sqrt{66}}{2*3}=\frac{-24+4\sqrt{66}}{6}
$