Let  $f\left(x\right)=\{\begin{array}{cc}4x-2;& 0\le x\le 1\\ -x^2-2x;& -2\le x<0\end{array}$ the number of solution(s) of the equation  $f\left(f\left(x\right)\right)=\frac{3}{4}$ is_______

$$\begin{gathered} f\left(x\right)=t\Rightarrow f\left(t\right)=\frac{3}{4} \\ \Rightarrow 4t-2=\frac{3}{4}, 1-t^2-2x=\frac{3}{4} \\ \Rightarrow t=\frac{11}{16}, t^2+2t+1=1-\frac{3}{4}=\frac{1}{4} \\ \Rightarrow {(t+1)}^2=\frac{1}{4} \Rightarrow t=\frac{-1}{2},\frac{-3}{2} \\ f\left(x\right)=\frac{11}{16}\to 3 solutions \\ f\left(x\right)=\frac{-1}{2}\to 1 solution \\ f\left(x\right)=\frac{-3}{2}\to 1 solution \Rightarrow 5 solutions are possible \end{gathered}$$