1^2+x^2=32^2

Solution for 1^2+x^2=32^2 equation:

1^2+x^2=32^2

We move all terms to the left:

1^2+x^2-(32^2)=0

We add all the numbers together, and all the variables

x^2-1023=0

a = 1; b = 0; c = -1023;
Δ = b2-4ac
Δ = 02-4·1·(-1023)
Δ = 4092
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:

$\sqrt{\Delta}=\sqrt{4092}=\sqrt{4*1023}=\sqrt{4}*\sqrt{1023}=2\sqrt{1023}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{1023}}{2*1}=\frac{0-2\sqrt{1023}}{2}
=-\frac{2\sqrt{1023}}{2}
=-\sqrt{1023}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{1023}}{2*1}=\frac{0+2\sqrt{1023}}{2}
=\frac{2\sqrt{1023}}{2}
=\sqrt{1023}
$