2x^2-22=16x

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

2x^2-22=16x

We move all terms to the left:

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

a = 2; b = -16; c = -22;
Δ = b2-4ac
Δ = -162-4·2·(-22)
Δ = 432
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:

$\sqrt{\Delta}=\sqrt{432}=\sqrt{144*3}=\sqrt{144}*\sqrt{3}=12\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-12\sqrt{3}}{2*2}=\frac{16-12\sqrt{3}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+12\sqrt{3}}{2*2}=\frac{16+12\sqrt{3}}{4}
$