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

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

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

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

v^2-3v-28=0

a = 1; b = -3; c = -28;
Δ = b2-4ac
Δ = -32-4·1·(-28)
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*1}=\frac{-8}{2}
=-4
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+11}{2*1}=\frac{14}{2}
=7
$