(6x)^2+(2x)^2=40000

Solution for (6x)^2+(2x)^2=40000 equation:

(6x)^2+(2x)^2=40000

We move all terms to the left:

(6x)^2+(2x)^2-(40000)=0

We add all the numbers together, and all the variables

8x^2-40000=0

a = 8; b = 0; c = -40000;
Δ = b2-4ac
Δ = 02-4·8·(-40000)
Δ = 1280000
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:

$\sqrt{\Delta}=\sqrt{1280000}=\sqrt{640000*2}=\sqrt{640000}*\sqrt{2}=800\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-800\sqrt{2}}{2*8}=\frac{0-800\sqrt{2}}{16}
=-\frac{800\sqrt{2}}{16}
=-50\sqrt{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+800\sqrt{2}}{2*8}=\frac{0+800\sqrt{2}}{16}
=\frac{800\sqrt{2}}{16}
=50\sqrt{2}
$