-16x^2+200x-612=0

Solution for -16x^2+200x-612=0 equation:

-16x^2+200x-612=0

a = -16; b = 200; c = -612;
Δ = b2-4ac
Δ = 2002-4·(-16)·(-612)
Δ = 832
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:

$\sqrt{\Delta}=\sqrt{832}=\sqrt{64*13}=\sqrt{64}*\sqrt{13}=8\sqrt{13}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(200)-8\sqrt{13}}{2*-16}=\frac{-200-8\sqrt{13}}{-32}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(200)+8\sqrt{13}}{2*-16}=\frac{-200+8\sqrt{13}}{-32}
$