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

Solution for (2v-7)(4+v)=0 equation:

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(+2v^2+8v-7v-28)=0

We get rid of parentheses

2v^2+8v-7v-28=0

We add all the numbers together, and all the variables

2v^2+v-28=0

a = 2; b = 1; c = -28;
Δ = b2-4ac
Δ = 12-4·2·(-28)
Δ = 225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{225}=15$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-15}{2*2}=\frac{-16}{4}
=-4
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+15}{2*2}=\frac{14}{4}
=3+1/2
$