-16x^2+96x+50=0

Solution for -16x^2+96x+50=0 equation:

-16x^2+96x+50=0

a = -16; b = 96; c = +50;
Δ = b2-4ac
Δ = 962-4·(-16)·50
Δ = 12416
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:

$\sqrt{\Delta}=\sqrt{12416}=\sqrt{64*194}=\sqrt{64}*\sqrt{194}=8\sqrt{194}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(96)-8\sqrt{194}}{2*-16}=\frac{-96-8\sqrt{194}}{-32}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(96)+8\sqrt{194}}{2*-16}=\frac{-96+8\sqrt{194}}{-32}
$