Solution
= Solution
A real square on the left requires $x^5\geq0$, so $x\geq0$. Taking the two square roots gives exactly the two real graph branches
$$
\boxed{y_+(x)=x^2+x^{5/2},\qquad y_-(x)=x^2-x^{5/2},\qquad x\geq0}.
$$
They meet only at the origin. The local coordinate change $Y=y-x^2$ turns the equation into $Y^2=x^5$, the <plane cusp of type (2,5)>; both branches share the tangent $y=0$ there.