Solve v+5=-2(x-1)^2

$$v + 5 = - 2 ( x - 1 ) ^ { 2 }$$

Solve for v

$$v=-2\left(x-1\right)^{2}-5$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-1\right)^{2}.

$$v+5=-2\left(x^{2}-2x+1\right)$$

Use the distributive property to multiply -2 by x^{2}-2x+1.

$$v+5=-2x^{2}+4x-2$$

Subtract 5 from both sides.

$$v=-2x^{2}+4x-2-5$$

Subtract 5 from -2 to get -7.

$$v=-2x^{2}+4x-7$$

Solve for x (complex solution)

$$x=-\frac{\sqrt{-2v-10}}{2}+1$$

$$x=\frac{\sqrt{-2v-10}}{2}+1$$

Solve for x

$$x=-\frac{\sqrt{-2v-10}}{2}+1$$

$$x=\frac{\sqrt{-2v-10}}{2}+1\text{, }v\leq -5$$