(1+v)(4v-5)=0

Solution for (1+v)(4v-5)=0 equation:

(1+v)(4v-5)=0

We add all the numbers together, and all the variables

(v+1)(4v-5)=0

We multiply parentheses ..

(+4v^2-5v+4v-5)=0

We get rid of parentheses

4v^2-5v+4v-5=0

We add all the numbers together, and all the variables

4v^2-1v-5=0

a = 4; b = -1; c = -5;
Δ = b2-4ac
Δ = -12-4·4·(-5)
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-9}{2*4}=\frac{-8}{8}
=-1
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+9}{2*4}=\frac{10}{8}
=1+1/4
$