2x^2+3x^2=208

Solution for 2x^2+3x^2=208 equation:

2x^2+3x^2=208

We move all terms to the left:

2x^2+3x^2-(208)=0

We add all the numbers together, and all the variables

5x^2-208=0

a = 5; b = 0; c = -208;
Δ = b2-4ac
Δ = 02-4·5·(-208)
Δ = 4160
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:

$\sqrt{\Delta}=\sqrt{4160}=\sqrt{64*65}=\sqrt{64}*\sqrt{65}=8\sqrt{65}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{65}}{2*5}=\frac{0-8\sqrt{65}}{10}
=-\frac{8\sqrt{65}}{10}
=-\frac{4\sqrt{65}}{5}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{65}}{2*5}=\frac{0+8\sqrt{65}}{10}
=\frac{8\sqrt{65}}{10}
=\frac{4\sqrt{65}}{5}
$