(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+11v+7=0

a = 4; b = 11; c = +7;
Δ = b2-4ac
Δ = 112-4·4·7
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-3}{2*4}=\frac{-14}{8}
=-1+3/4
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+3}{2*4}=\frac{-8}{8}
=-1
$