How do you solve the equation to where the function $f(x)=1+\sqrt{x}$ and it's inverse $g(x)=(x-1)^2$ intersect?

The problem stated in the title comes from an entry-level university/upper secondary school textbook.

My initial attempt to solve it was to expand $(x-1)^2=x^2-2x+1$, from there $x^2-2x+1=1+\sqrt{x}$ and I can subtract 1 from each side. From there I square both sides which gives me $x=(x^2-2x)^2$. If expand the right hand side I get $(x^2-2x)^2=x^4-4x^3+4x^2$. Then I go back and subtract x from both sides and I get $x^4-4x^3+4x^2-x=0$. From this point on however, I have trouble finding a way to continue. I can see how I can divide the terms with x but I don't know if that would help. Since there are the terms $4x^3$ and $-x$ I also can't express $x^4$ and $4x^2$ as $t^2$ and $t$ and solve it using the quadratic formula. The goal is to find the exact point at which these functions intersect. Possibly there is some easier solution than solving it this way so I would be happy to hear any advice. Thanks in advance!

Note that, a strictly increasing function, always intersects its inverse on $y=x$ line. Hence it is enough to find intersection points of $f(x)=1+\sqrt x$ and the line $y=x$,

$$1+\sqrt x=x\Rightarrow (\sqrt x)^2-\sqrt x-1=0$$ $$\sqrt x=\frac{1\pm\sqrt5}{2}$$ Since $\sqrt x\ge0$ , only $\sqrt x=\frac{1+\sqrt5}2$ is acceptable. Hence $x=\left(\frac{1+\sqrt5}2\right)^2=\frac{3+\sqrt5}2$ and intersection point is $\large(\frac{3+\sqrt5}2,\frac{3+\sqrt5}2)$.