-16x^2+16x=-200

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

-16x^2+16x=-200

We move all terms to the left:

-16x^2+16x-(-200)=0

We add all the numbers together, and all the variables

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

a = -16; b = 16; c = +200;
Δ = b2-4ac
Δ = 162-4·(-16)·200
Δ = 13056
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:

$\sqrt{\Delta}=\sqrt{13056}=\sqrt{256*51}=\sqrt{256}*\sqrt{51}=16\sqrt{51}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-16\sqrt{51}}{2*-16}=\frac{-16-16\sqrt{51}}{-32}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+16\sqrt{51}}{2*-16}=\frac{-16+16\sqrt{51}}{-32}
$