-16x^2+40x+15=0

Solution for -16x^2+40x+15=0 equation:

-16x^2+40x+15=0

a = -16; b = 40; c = +15;
Δ = b2-4ac
Δ = 402-4·(-16)·15
Δ = 2560
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:

$\sqrt{\Delta}=\sqrt{2560}=\sqrt{256*10}=\sqrt{256}*\sqrt{10}=16\sqrt{10}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-16\sqrt{10}}{2*-16}=\frac{-40-16\sqrt{10}}{-32}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+16\sqrt{10}}{2*-16}=\frac{-40+16\sqrt{10}}{-32}
$