-16x2+32=-128

Solution for -16x2+32=-128 equation:

-16x^2+32=-128

We move all terms to the left:

-16x^2+32-(-128)=0

We add all the numbers together, and all the variables

-16x^2+160=0

a = -16; b = 0; c = +160;
Δ = b2-4ac
Δ = 02-4·(-16)·160
Δ = 10240
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:

$\sqrt{\Delta}=\sqrt{10240}=\sqrt{1024*10}=\sqrt{1024}*\sqrt{10}=32\sqrt{10}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-32\sqrt{10}}{2*-16}=\frac{0-32\sqrt{10}}{-32}
=-\frac{32\sqrt{10}}{-32}
=-\frac{\sqrt{10}}{-1}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+32\sqrt{10}}{2*-16}=\frac{0+32\sqrt{10}}{-32}
=\frac{32\sqrt{10}}{-32}
=\frac{\sqrt{10}}{-1}
$