Question

system of equations (pre-calculus)

Answer 1

\[y=\sqrt{324-x^2}\]
\[y=20-x\]

Answer 2

so i substitued 20-x for Y in the first equation so i have this
\[\sqrt{324-x^2}=20-x\]

Answer 3

I was told i square both sides and i should get
\[324-x^2=400-40x+x^2\]
but I don't understand where the 40 comes from. I'd appreciate any help. thanks!