Solve x=(y+1)^2+8

$$x = ( y + 1 ) ^ { 2 } + 8$$

Solve for x

$$x=\left(y+1\right)^{2}+8$$

Solution Steps

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

$$x=y^{2}+2y+1+8$$

Add 1 and 8 to get 9.

$$x=y^{2}+2y+9$$

Solve for y (complex solution)

$$y=-\left(\sqrt{x-8}+1\right)$$

$$y=\sqrt{x-8}-1$$

Solve for y

$$y=-\left(\sqrt{x-8}+1\right)$$

$$y=\sqrt{x-8}-1\text{, }x\geq 8$$