-16x^2+32x=-6

Solution for -16x^2+32x=-6 equation:

-16x^2+32x=-6

We move all terms to the left:

-16x^2+32x-(-6)=0

We add all the numbers together, and all the variables

-16x^2+32x+6=0

a = -16; b = 32; c = +6;
Δ = b2-4ac
Δ = 322-4·(-16)·6
Δ = 1408
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:

$\sqrt{\Delta}=\sqrt{1408}=\sqrt{64*22}=\sqrt{64}*\sqrt{22}=8\sqrt{22}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-8\sqrt{22}}{2*-16}=\frac{-32-8\sqrt{22}}{-32}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+8\sqrt{22}}{2*-16}=\frac{-32+8\sqrt{22}}{-32}
$