x^2+(2x)^2=2.25

Solution for x^2+(2x)^2=2.25 equation:

x^2+(2x)^2=2.25

We move all terms to the left:

x^2+(2x)^2-(2.25)=0

We add all the numbers together, and all the variables

3x^2-2.25=0

a = 3; b = 0; c = -2.25;
Δ = b2-4ac
Δ = 02-4·3·(-2.25)
Δ = 27
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:

$\sqrt{\Delta}=\sqrt{27}=\sqrt{9*3}=\sqrt{9}*\sqrt{3}=3\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-3\sqrt{3}}{2*3}=\frac{0-3\sqrt{3}}{6}
=-\frac{3\sqrt{3}}{6}
=-\frac{\sqrt{3}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+3\sqrt{3}}{2*3}=\frac{0+3\sqrt{3}}{6}
=\frac{3\sqrt{3}}{6}
=\frac{\sqrt{3}}{2}
$