(v^2+6v-7)/(3v^2-3)=0

Solution for (v^2+6v-7)/(3v^2-3)=0 equation:

(v^2+6v-7)/(3v^2-3)=0


Domain of the equation: (3v^2-3)!=0

We move all terms containing v to the left, all other terms to the right

3v^2!=3

v^2!=3/3

v^2!=√1

v!=1

v∈R


We multiply all the terms by the denominator


(v^2+6v-7)=0

We get rid of parentheses

v^2+6v-7=0

a = 1; b = 6; c = -7;
Δ = b2-4ac
Δ = 62-4·1·(-7)
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-8}{2*1}=\frac{-14}{2}
=-7
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+8}{2*1}=\frac{2}{2}
=1
$