16^2+x^2=27^2

Solution for 16^2+x^2=27^2 equation:

16^2+x^2=27^2

We move all terms to the left:

16^2+x^2-(27^2)=0

We add all the numbers together, and all the variables

x^2-473=0

a = 1; b = 0; c = -473;
Δ = b2-4ac
Δ = 02-4·1·(-473)
Δ = 1892
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:

$\sqrt{\Delta}=\sqrt{1892}=\sqrt{4*473}=\sqrt{4}*\sqrt{473}=2\sqrt{473}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{473}}{2*1}=\frac{0-2\sqrt{473}}{2}
=-\frac{2\sqrt{473}}{2}
=-\sqrt{473}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{473}}{2*1}=\frac{0+2\sqrt{473}}{2}
=\frac{2\sqrt{473}}{2}
=\sqrt{473}
$