10^2+(2x)^2=500

Solution for 10^2+(2x)^2=500 equation:

10^2+(2x)^2=500

We move all terms to the left:

10^2+(2x)^2-(500)=0

determiningTheFunctionDomain
2x^2-500+10^2=0

We add all the numbers together, and all the variables

2x^2-400=0

a = 2; b = 0; c = -400;
Δ = b2-4ac
Δ = 02-4·2·(-400)
Δ = 3200
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:

$\sqrt{\Delta}=\sqrt{3200}=\sqrt{1600*2}=\sqrt{1600}*\sqrt{2}=40\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-40\sqrt{2}}{2*2}=\frac{0-40\sqrt{2}}{4}
=-\frac{40\sqrt{2}}{4}
=-10\sqrt{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+40\sqrt{2}}{2*2}=\frac{0+40\sqrt{2}}{4}
=\frac{40\sqrt{2}}{4}
=10\sqrt{2}
$