(4v+3)(1+v)=0

Solution for (4v+3)(1+v)=0 equation:

(4v+3)(1+v)=0

We add all the numbers together, and all the variables

(4v+3)(v+1)=0

We multiply parentheses ..

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

We get rid of parentheses

4v^2+4v+3v+3=0

We add all the numbers together, and all the variables

4v^2+7v+3=0

a = 4; b = 7; c = +3;
Δ = b2-4ac
Δ = 72-4·4·3
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-1}{2*4}=\frac{-8}{8}
=-1
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+1}{2*4}=\frac{-6}{8}
=-3/4
$