x^2-22x+16=90

Solution for x^2-22x+16=90 equation:

x^2-22x+16=90

We move all terms to the left:

x^2-22x+16-(90)=0

We add all the numbers together, and all the variables

x^2-22x-74=0

a = 1; b = -22; c = -74;
Δ = b2-4ac
Δ = -222-4·1·(-74)
Δ = 780
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:

$\sqrt{\Delta}=\sqrt{780}=\sqrt{4*195}=\sqrt{4}*\sqrt{195}=2\sqrt{195}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-22)-2\sqrt{195}}{2*1}=\frac{22-2\sqrt{195}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-22)+2\sqrt{195}}{2*1}=\frac{22+2\sqrt{195}}{2}
$