Solve v+2=-(x+4)^2

$$v + 2 = - ( x + 4 ) ^ { 2 }$$

Solve for v

$$v=-\left(\left(x+4\right)^{2}+2\right)$$

Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+4\right)^{2}.

$$v+2=-\left(x^{2}+8x+16\right)$$

To find the opposite of x^{2}+8x+16, find the opposite of each term.

$$v+2=-x^{2}-8x-16$$

Subtract 2 from both sides.

$$v=-x^{2}-8x-16-2$$

Subtract 2 from -16 to get -18.

$$v=-x^{2}-8x-18$$

Solve for x (complex solution)

$$x=-i\sqrt{v+2}-4$$

$$x=i\sqrt{v+2}-4$$

Solve for x

$$x=-\left(\sqrt{-v-2}+4\right)$$

$$x=\sqrt{-v-2}-4\text{, }v\leq -2$$