1600=(2x)^2+(3x)^2

Solution for 1600=(2x)^2+(3x)^2 equation:

1600=(2x)^2+(3x)^2

We move all terms to the left:

1600-((2x)^2+(3x)^2)=0

We get rid of parentheses

-2x^2-3x^2+1600=0

We add all the numbers together, and all the variables

-5x^2+1600=0

a = -5; b = 0; c = +1600;
Δ = b2-4ac
Δ = 02-4·(-5)·1600
Δ = 32000
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:

$\sqrt{\Delta}=\sqrt{32000}=\sqrt{6400*5}=\sqrt{6400}*\sqrt{5}=80\sqrt{5}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-80\sqrt{5}}{2*-5}=\frac{0-80\sqrt{5}}{-10}
=-\frac{80\sqrt{5}}{-10}
=-\frac{8\sqrt{5}}{-1}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+80\sqrt{5}}{2*-5}=\frac{0+80\sqrt{5}}{-10}
=\frac{80\sqrt{5}}{-10}
=\frac{8\sqrt{5}}{-1}
$