16^2+x^2=22^2

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

16^2+x^2=22^2

We move all terms to the left:

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

We add all the numbers together, and all the variables

x^2-228=0

a = 1; b = 0; c = -228;
Δ = b2-4ac
Δ = 02-4·1·(-228)
Δ = 912
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:

$\sqrt{\Delta}=\sqrt{912}=\sqrt{16*57}=\sqrt{16}*\sqrt{57}=4\sqrt{57}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{57}}{2*1}=\frac{0-4\sqrt{57}}{2}
=-\frac{4\sqrt{57}}{2}
=-2\sqrt{57}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{57}}{2*1}=\frac{0+4\sqrt{57}}{2}
=\frac{4\sqrt{57}}{2}
=2\sqrt{57}
$