210.5^2+x^2=421^2

Solution for 210.5^2+x^2=421^2 equation:

210.5^2+x^2=421^2

We move all terms to the left:

210.5^2+x^2-(421^2)=0

We add all the numbers together, and all the variables

x^2-132930.75=0

a = 1; b = 0; c = -132930.75;
Δ = b2-4ac
Δ = 02-4·1·(-132930.75)
Δ = 531723
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:

$\sqrt{\Delta}=\sqrt{531723}=\sqrt{177241*3}=\sqrt{177241}*\sqrt{3}=421\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-421\sqrt{3}}{2*1}=\frac{0-421\sqrt{3}}{2}
=-\frac{421\sqrt{3}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+421\sqrt{3}}{2*1}=\frac{0+421\sqrt{3}}{2}
=\frac{421\sqrt{3}}{2}
$