Solve for v v^2+4v+9=-2v

$v^{2} + 4 v + 9 = - 2 v$

Move all terms containing $v$ to the left side of the equation.

Add $2 v$ to both sides of the equation.

$v^{2} + 4 v + 9 + 2 v = 0$

Add $4 v$ and $2 v$ .

$v^{2} + 6 v + 9 = 0$

Factor using the perfect square rule.

Rewrite $9$ as $3^{2}$ .

$v^{2} + 6 v + 3^{2} = 0$

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$6 v = 2 \cdot v \cdot 3$

Rewrite the polynomial.

$v^{2} + 2 \cdot v \cdot 3 + 3^{2} = 0$

Factor using the perfect square trinomial rule $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ , where $a = v$ and $b = 3$ .

${(v + 3)}^{2} = 0$

Set the $v + 3$ equal to $0$ .

$v + 3 = 0$

Subtract $3$ from both sides of the equation.

$v = - 3$