2v^2+3v^2=1

Solution for 2v^2+3v^2=1 equation:

2v^2+3v^2=1

We move all terms to the left:

2v^2+3v^2-(1)=0

We add all the numbers together, and all the variables

5v^2-1=0

a = 5; b = 0; c = -1;
Δ = b2-4ac
Δ = 02-4·5·(-1)
Δ = 20
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{20}=\sqrt{4*5}=\sqrt{4}*\sqrt{5}=2\sqrt{5}$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{5}}{2*5}=\frac{0-2\sqrt{5}}{10}
=-\frac{2\sqrt{5}}{10}
=-\frac{\sqrt{5}}{5}
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{5}}{2*5}=\frac{0+2\sqrt{5}}{10}
=\frac{2\sqrt{5}}{10}
=\frac{\sqrt{5}}{5}
$