Solve x^2-10+18=(x-a)^2+b

$$x ^ { 2 } - 10 + 18 = ( x - a ) ^ { 2 } + b$$

Solve for b

$$b=a\left(2x-a\right)+8$$

Solution Steps

Add -10 and 18 to get 8.

$$x^{2}+8=\left(x-a\right)^{2}+b$$

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

$$x^{2}+8=x^{2}-2xa+a^{2}+b$$

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

$$x^{2}-2xa+a^{2}+b=x^{2}+8$$

Subtract x^{2} from both sides.

$$-2xa+a^{2}+b=x^{2}+8-x^{2}$$

Combine x^{2} and -x^{2} to get 0.

$$-2xa+a^{2}+b=8$$

Add 2xa to both sides.

$$a^{2}+b=8+2xa$$

Subtract a^{2} from both sides.

$$b=8+2xa-a^{2}$$

Solve for a (complex solution)

$$a=\sqrt{x^{2}-b+8}+x$$

$$a=-\sqrt{x^{2}-b+8}+x$$

Solve for a

$$a=\sqrt{x^{2}-b+8}+x$$

$$a=-\sqrt{x^{2}-b+8}+x\text{, }b\leq x^{2}+8$$