-2x^2+50x-128=0

Solution for -2x^2+50x-128=0 equation:

-2x^2+50x-128=0

a = -2; b = 50; c = -128;
Δ = b2-4ac
Δ = 502-4·(-2)·(-128)
Δ = 1476
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:

$\sqrt{\Delta}=\sqrt{1476}=\sqrt{36*41}=\sqrt{36}*\sqrt{41}=6\sqrt{41}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(50)-6\sqrt{41}}{2*-2}=\frac{-50-6\sqrt{41}}{-4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(50)+6\sqrt{41}}{2*-2}=\frac{-50+6\sqrt{41}}{-4}
$