(v^2+2v-9)/(v+1)^2=0

Solution for (v^2+2v-9)/(v+1)^2=0 equation:

(v^2+2v-9)/(v+1)^2=0


Domain of the equation: (v+1)^2!=0

v∈R


We multiply all the terms by the denominator


(v^2+2v-9)=0

We get rid of parentheses

v^2+2v-9=0

a = 1; b = 2; c = -9;
Δ = b2-4ac
Δ = 22-4·1·(-9)
Δ = 40
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{40}=\sqrt{4*10}=\sqrt{4}*\sqrt{10}=2\sqrt{10}$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{10}}{2*1}=\frac{-2-2\sqrt{10}}{2}
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{10}}{2*1}=\frac{-2+2\sqrt{10}}{2}
$