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

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

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

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

4v^2+v-3=0

a = 4; b = 1; c = -3;
Δ = b2-4ac
Δ = 12-4·4·(-3)
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-7}{2*4}=\frac{-8}{8}
=-1
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+7}{2*4}=\frac{6}{8}
=3/4
$