Solve y=(x+p)^2+q

$$y = ( x + p ) ^ { 2 } + q$$

Solve for q

$$q=-\left(x+p\right)^{2}+y$$

Solution Steps

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

$$y=x^{2}+2xp+p^{2}+q$$

Swap sides so that all variable terms are on the left hand side.

$$x^{2}+2xp+p^{2}+q=y$$

Subtract x^{2} from both sides.

$$2xp+p^{2}+q=y-x^{2}$$

Subtract 2xp from both sides.

$$p^{2}+q=y-x^{2}-2xp$$

Subtract p^{2} from both sides.

$$q=y-x^{2}-2xp-p^{2}$$

Solve for p (complex solution)

$$p=-\left(x+\sqrt{y-q}\right)$$

$$p=-x+\sqrt{y-q}$$

Solve for p

$$p=-\left(x+\sqrt{y-q}\right)$$

$$p=-x+\sqrt{y-q}\text{, }y\geq q$$