4x^2+(4x)^2=20

Solution for 4x^2+(4x)^2=20 equation:

4x^2+(4x)^2=20

We move all terms to the left:

4x^2+(4x)^2-(20)=0

We add all the numbers together, and all the variables

8x^2-20=0

a = 8; b = 0; c = -20;
Δ = b2-4ac
Δ = 02-4·8·(-20)
Δ = 640
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:

$\sqrt{\Delta}=\sqrt{640}=\sqrt{64*10}=\sqrt{64}*\sqrt{10}=8\sqrt{10}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{10}}{2*8}=\frac{0-8\sqrt{10}}{16}
=-\frac{8\sqrt{10}}{16}
=-\frac{\sqrt{10}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{10}}{2*8}=\frac{0+8\sqrt{10}}{16}
=\frac{8\sqrt{10}}{16}
=\frac{\sqrt{10}}{2}
$