-8x^2+40x+15=0

Solution for -8x^2+40x+15=0 equation:

-8x^2+40x+15=0

a = -8; b = 40; c = +15;
Δ = b2-4ac
Δ = 402-4·(-8)·15
Δ = 2080
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:

$\sqrt{\Delta}=\sqrt{2080}=\sqrt{16*130}=\sqrt{16}*\sqrt{130}=4\sqrt{130}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-4\sqrt{130}}{2*-8}=\frac{-40-4\sqrt{130}}{-16}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+4\sqrt{130}}{2*-8}=\frac{-40+4\sqrt{130}}{-16}
$