Solve y=left(x-3right)^2-5

$$y= { \left(x-3 \right) }^{ 2 } -5$$

Solve for y

$$y=\left(x-3\right)^{2}-5$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-3\right)^{2}.

$$y=x^{2}-6x+9-5$$

Subtract 5 from 9 to get 4.

$$y=x^{2}-6x+4$$

Solve for x (complex solution)

$$x=-\sqrt{y+5}+3$$

$$x=\sqrt{y+5}+3$$

Solve for x

$$x=-\sqrt{y+5}+3$$

$$x=\sqrt{y+5}+3\text{, }y\geq -5$$