-16x^2+32x+20=x

Solution for -16x^2+32x+20=x equation:

-16x^2+32x+20=x

We move all terms to the left:

-16x^2+32x+20-(x)=0

We add all the numbers together, and all the variables

-16x^2+31x+20=0

a = -16; b = 31; c = +20;
Δ = b2-4ac
Δ = 312-4·(-16)·20
Δ = 2241
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:

$\sqrt{\Delta}=\sqrt{2241}=\sqrt{9*249}=\sqrt{9}*\sqrt{249}=3\sqrt{249}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(31)-3\sqrt{249}}{2*-16}=\frac{-31-3\sqrt{249}}{-32}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(31)+3\sqrt{249}}{2*-16}=\frac{-31+3\sqrt{249}}{-32}
$