Proof showing if $f(\frac{3x-2}{2})=x^2-x-1,$ find $f(0)=?$

Question

PS: Before posting it, one tried to grasp this (although didn't understand mfl's answer completely either).

I was shown the way to solve it: if I set $$\frac{3x-2}{2}=0 \Rightarrow x=\frac{2}{3} \Rightarrow x^2-x-1=-\frac{11}{9}$$ that answer will also be the one of $f(0)$ (i.e. $f(0)=-11/9$), which is true, but I don't quite get it: how the answer of $f(\frac{3x-2}{2})$ (i.e. $-11/9$) is also the answer of $f(0)$? why the answer of the quad equ. is also the answer of $f(0)$ (that it is $f(0)=-11/9$)? I asked the person who solved it for me if there was a proof for this, but she couldn't bring it. So, I would be very happy and obliged if someone here would show the proof that it is $f(0)=-11/9$.

Answer 1

substituting $$t=\frac{3x-2}{3}$$ then we get $$x=\frac{2t+2}{3}$$ then we get $$f(t)=\left(\frac{2t+2}{3}\right)^2-\frac{2+2t}{3}-1$$ then we get ...?

Answer 2

Hint: To calculate $f(0)$ given an expression for $f((3x-2)/2)$, you could find the value of $x$ that makes the argument zero, and then use that value in the expression.