(4v-7)(v+1)=0

Solution for (4v-7)(v+1)=0 equation:

(4v-7)(v+1)=0

We multiply parentheses ..

(+4v^2+4v-7v-7)=0

We get rid of parentheses

4v^2+4v-7v-7=0

We add all the numbers together, and all the variables

4v^2-3v-7=0

a = 4; b = -3; c = -7;
Δ = b2-4ac
Δ = -32-4·4·(-7)
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-11}{2*4}=\frac{-8}{8}
=-1
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+11}{2*4}=\frac{14}{8}
=1+3/4
$