-16x^2-22x+5=2x

Solution for -16x^2-22x+5=2x equation:

-16x^2-22x+5=2x

We move all terms to the left:

-16x^2-22x+5-(2x)=0

We add all the numbers together, and all the variables

-16x^2-24x+5=0

a = -16; b = -24; c = +5;
Δ = b2-4ac
Δ = -242-4·(-16)·5
Δ = 896
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:

$\sqrt{\Delta}=\sqrt{896}=\sqrt{64*14}=\sqrt{64}*\sqrt{14}=8\sqrt{14}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-24)-8\sqrt{14}}{2*-16}=\frac{24-8\sqrt{14}}{-32}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-24)+8\sqrt{14}}{2*-16}=\frac{24+8\sqrt{14}}{-32}
$