-3x^2-40x-5=0

Solution for -3x^2-40x-5=0 equation:

-3x^2-40x-5=0

a = -3; b = -40; c = -5;
Δ = b2-4ac
Δ = -402-4·(-3)·(-5)
Δ = 1540
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:

$\sqrt{\Delta}=\sqrt{1540}=\sqrt{4*385}=\sqrt{4}*\sqrt{385}=2\sqrt{385}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-2\sqrt{385}}{2*-3}=\frac{40-2\sqrt{385}}{-6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+2\sqrt{385}}{2*-3}=\frac{40+2\sqrt{385}}{-6}
$