x^2+(2x)^2=1.96

Solution for x^2+(2x)^2=1.96 equation:

x^2+(2x)^2=1.96

We move all terms to the left:

x^2+(2x)^2-(1.96)=0

We add all the numbers together, and all the variables

3x^2-1.96=0

a = 3; b = 0; c = -1.96;
Δ = b2-4ac
Δ = 02-4·3·(-1.96)
Δ = 23.52
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-\sqrt{23.52}}{2*3}=\frac{0-\sqrt{23.52}}{6}
=-\frac{\sqrt{}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+\sqrt{23.52}}{2*3}=\frac{0+\sqrt{23.52}}{6}
=\frac{\sqrt{}}{6}
$