0=v^2+4v+3

Solution for 0=v^2+4v+3 equation:

0=v^2+4v+3

We move all terms to the left:

0-(v^2+4v+3)=0

We add all the numbers together, and all the variables

-(v^2+4v+3)=0

We get rid of parentheses

-v^2-4v-3=0

We add all the numbers together, and all the variables

-1v^2-4v-3=0

a = -1; b = -4; c = -3;
Δ = b2-4ac
Δ = -42-4·(-1)·(-3)
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-2}{2*-1}=\frac{2}{-2}
=-1
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+2}{2*-1}=\frac{6}{-2}
=-3
$