(5v-7)(4+v)=0

Solution for (5v-7)(4+v)=0 equation:

(5v-7)(4+v)=0

We add all the numbers together, and all the variables

(5v-7)(v+4)=0

We multiply parentheses ..

(+5v^2+20v-7v-28)=0

We get rid of parentheses

5v^2+20v-7v-28=0

We add all the numbers together, and all the variables

5v^2+13v-28=0

a = 5; b = 13; c = -28;
Δ = b2-4ac
Δ = 132-4·5·(-28)
Δ = 729
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{729}=27$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-27}{2*5}=\frac{-40}{10}
=-4
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+27}{2*5}=\frac{14}{10}
=1+2/5
$