2x^2+x^2=4096

Solution for 2x^2+x^2=4096 equation:

2x^2+x^2=4096

We move all terms to the left:

2x^2+x^2-(4096)=0

We add all the numbers together, and all the variables

3x^2-4096=0

a = 3; b = 0; c = -4096;
Δ = b2-4ac
Δ = 02-4·3·(-4096)
Δ = 49152
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:

$\sqrt{\Delta}=\sqrt{49152}=\sqrt{16384*3}=\sqrt{16384}*\sqrt{3}=128\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-128\sqrt{3}}{2*3}=\frac{0-128\sqrt{3}}{6}
=-\frac{128\sqrt{3}}{6}
=-\frac{64\sqrt{3}}{3}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+128\sqrt{3}}{2*3}=\frac{0+128\sqrt{3}}{6}
=\frac{128\sqrt{3}}{6}
=\frac{64\sqrt{3}}{3}
$