Solve for y
y=\left(x-3\right)^{2}+1
Solve for x (complex solution)
x=-\sqrt{y-1}+3
x=\sqrt{y-1}+3
Solve for x
x=-\sqrt{y-1}+3
x=\sqrt{y-1}+3\text{, }y\geq 1
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y=x^{2}-6x+9+1
Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-3\right)^{2}.
y=x^{2}-6x+10
Add 9 and 1 to get 10.
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