-16x^2+320x+32=x

Solution for -16x^2+320x+32=x equation:

-16x^2+320x+32=x

We move all terms to the left:

-16x^2+320x+32-(x)=0

We add all the numbers together, and all the variables

-16x^2+319x+32=0

a = -16; b = 319; c = +32;
Δ = b2-4ac
Δ = 3192-4·(-16)·32
Δ = 103809
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(319)-\sqrt{103809}}{2*-16}=\frac{-319-\sqrt{103809}}{-32}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(319)+\sqrt{103809}}{2*-16}=\frac{-319+\sqrt{103809}}{-32}
$