2v^2+4v+1=v^2+6v+9

Solution for 2v^2+4v+1=v^2+6v+9 equation:

2v^2+4v+1=v^2+6v+9

We move all terms to the left:

2v^2+4v+1-(v^2+6v+9)=0

We get rid of parentheses

2v^2-v^2+4v-6v-9+1=0

We add all the numbers together, and all the variables

v^2-2v-8=0

a = 1; b = -2; c = -8;
Δ = b2-4ac
Δ = -22-4·1·(-8)
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-6}{2*1}=\frac{-4}{2}
=-2
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+6}{2*1}=\frac{8}{2}
=4
$