Решить уравнение: 9^x+6^x-4^(x+0.5)=0

[latex]9^x+6^x-4^{x+0.5}=0 \\ \\ 9^x+6^x-4^{0.5}*4^x=0 \\ \\ 9^x+6^x-2*4^x=0 \ | : 4^x \ \textgreater \ 0 \\ \\ \frac{9^x}{4^x}+\frac{6^x}{4^x}-2=0 \\ \\ (\frac{3^x}{2^x})^2 + \frac{3^x*2^x}{2^x*2^x}-2=0 \\ \\ ((\frac{3}{2})^x)^2+(\frac{3}{2})^x-2=0 \\ \\ \left[\begin{array}{c}(\frac{3}{2})^x = t \ \textgreater \ 0\end{array}\right] \\ \\ t^2+t-2=0; D = 1^2-4*1*(-2) = 9 = 3^2 \\ \\ t_1=\frac{-1+3}{2}=1; t_2=\frac{-1-3}{2} = -2 (t\ \textgreater \ 0 !) \\ \\ (\frac{3}{2})^x=1 =\ \textgreater \ (\frac{3}{2})^x = (\frac{3}{2}) ^ 0 =\ \textgreater \ x = 0 \\ \\ OTBET: 0.[/latex]