hard algebra...

Are you sure that wasn't a +5 on the end? 

\(f(x)=(2x^{3/2} -3x^{-3/2} )^2  -5\\ Find \;\;x\;\;when\;\;f(x)=5\\ (2x^{3/2} -3x^{-3/2} )^2  -5=5\\ (2x^{3/2} -3x^{-3/2} )^2  =10\\ 4x^3+9x^{-3}-12x^{0}=100\\ 4x^3+9x^{-3}-12x=100\\ 4x^6+9-12x^4=100x^3\\ 4x^6-12x^4-100x^3+9=0\\ \)

\(f(x)=(2x^{3/2} -3x^{-3/2} )^2  +5\\ Find \;\;x\;\;when\;\;f(x)=5\\ (2x^{3/2} -3x^{-3/2} )^2  +5=5\\ (2x^{3/2} -3x^{-3/2} )^2  =0\\ 4x^3+9x^{-3}-12x^{0}=0\\ 4x^3+9x^{-3}-12=0\\ 4x^6+9-12x^3=0\\ 4x^6-12x^3+9=0\\ let\;\;m=x^3\\ 4m^2-12m+9=0\\ (2m-3)^2=0\\ m=\pm\left[{\frac{3}{2}}\right]^{0.5}\\ x^3=\pm\left[{\frac{3}{2}}\right]\\ \text{But x is positive so}\\ x^3=\left[{\frac{3}{2}}\right]\\ x=\left[{\frac{3}{2}}\right]^{1/3}\\ \)

1.5^(1/3) = 1.1447142425533319

x= 1.14   (to 3 significant figures)

That it the first bit, the second bit does not look too difficult, give it a try yourself.  :)