v2+4v+3=

Solution for v2+4v+3= equation:

v2+4v+3=

We move all terms to the left:

v2+4v+3-()=0

We add all the numbers together, and all the variables

v^2+4v=0

a = 1; b = 4; c = 0;
Δ = b2-4ac
Δ = 42-4·1·0
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4}{2*1}=\frac{-8}{2}
=-4
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4}{2*1}=\frac{0}{2}
=0
$