\(h(x)=x+x^{ \frac{1}{2}}
\\h=x+ \sqrt{x}
\\x=t^2
\\h=t^2+t\)
Solve for t
\(t^2+t-h=0
\\
\\t= \frac{-1\pm \sqrt{1+4h} }{2}
\\x=t^2=(\frac{-1\pm \sqrt{1+4h} }{2} )^2\)
The inverse function is
\(h^{-1}(x)=(\frac{-1\pm \sqrt{1+4x} }{2} )^2\)
Now find \(h^{-1}(6)\)
\(h^{-1}(6)=(\frac{-1\pm \sqrt{1+4\times 6} }{2} )^2=(\frac{-1\pm 5 }{2} )^2
\\
\\h^{-1}(6)=4\text{ or }h^{-1}(6)=9\)