(v-5)(v+3)=-7

Solution for (v-5)(v+3)=-7 equation:

(v-5)(v+3)=-7

We move all terms to the left:

(v-5)(v+3)-(-7)=0

We add all the numbers together, and all the variables

(v-5)(v+3)+7=0

We multiply parentheses ..

(+v^2+3v-5v-15)+7=0

We get rid of parentheses

v^2+3v-5v-15+7=0

We add all the numbers together, and all the variables

v^2-2v-8=0

a = 1; b = -2; c = -8;
Δ = b2-4ac
Δ = -22-4·1·(-8)
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-6}{2*1}=\frac{-4}{2}
=-2
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+6}{2*1}=\frac{8}{2}
=4
$