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

Solution for 128+16x-16x^2=32(x) equation:

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

We move all terms to the left:

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

determiningTheFunctionDomain
-16x^2+16x-32x+128=0

We add all the numbers together, and all the variables

-16x^2-16x+128=0

a = -16; b = -16; c = +128;
Δ = b2-4ac
Δ = -162-4·(-16)·128
Δ = 8448
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:

$\sqrt{\Delta}=\sqrt{8448}=\sqrt{256*33}=\sqrt{256}*\sqrt{33}=16\sqrt{33}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-16\sqrt{33}}{2*-16}=\frac{16-16\sqrt{33}}{-32}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+16\sqrt{33}}{2*-16}=\frac{16+16\sqrt{33}}{-32}
$