(1+v)(4v-6)=0

Solution for (1+v)(4v-6)=0 equation:

(1+v)(4v-6)=0

We add all the numbers together, and all the variables

(v+1)(4v-6)=0

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

4v^2-2v-6=0

a = 4; b = -2; c = -6;
Δ = b2-4ac
Δ = -22-4·4·(-6)
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-10}{2*4}=\frac{-8}{8}
=-1
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+10}{2*4}=\frac{12}{8}
=1+1/2
$