3+2v^2=17v^2

Solution for 3+2v^2=17v^2 equation:

3+2v^2=17v^2

We move all terms to the left:

3+2v^2-(17v^2)=0

We add all the numbers together, and all the variables

-15v^2+3=0

a = -15; b = 0; c = +3;
Δ = b2-4ac
Δ = 02-4·(-15)·3
Δ = 180
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:

$\sqrt{\Delta}=\sqrt{180}=\sqrt{36*5}=\sqrt{36}*\sqrt{5}=6\sqrt{5}$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{5}}{2*-15}=\frac{0-6\sqrt{5}}{-30}
=-\frac{6\sqrt{5}}{-30}
=-\frac{\sqrt{5}}{-5}
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{5}}{2*-15}=\frac{0+6\sqrt{5}}{-30}
=\frac{6\sqrt{5}}{-30}
=\frac{\sqrt{5}}{-5}
$