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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(-4v^2+7v+16v-28)=0

We get rid of parentheses

-4v^2+7v+16v-28=0

We add all the numbers together, and all the variables

-4v^2+23v-28=0

a = -4; b = 23; c = -28;
Δ = b2-4ac
Δ = 232-4·(-4)·(-28)
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{-(23)-9}{2*-4}=\frac{-32}{-8}
=+4
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+9}{2*-4}=\frac{-14}{-8}
=1+3/4
$