Алгебра, решите пожалуйста

$\frac{ 3^{2x}-54* ( \frac{1}{3}) ^{2(x+1)}-1 }{x+3} \leq 0$

$\frac{ 3^{2x}-54* ( \frac{1}{3}) ^{2x+2}-1 }{x+3} \leq 0$

$\frac{ 3^{2x}-54* \frac{1}{9}*( \frac{1}{3}) ^{2x}-1 }{x+3} \leq 0$

$\frac{ 3^{2x}-6*( \frac{1}{3}) ^{2x}-1 }{x+3} \leq 0$

${ 3^{2x}-6*( \frac{1}{3}) ^{2x}-1 =0$

${ 3^{2x}=t,$ $t\ \textgreater \ 0$

$t- \frac{6}{t} -1=0$

$t^2-t-6=0$

$D=(-1)^2-4*1*(-6)=25$

$t_1= \frac{1+5}{2}=3$

$t_2= \frac{1-5}{2} =-2$

$\frac{ ( 3^{2x}-3)( 3^{2x}+2) }{x+3} \leq 0$

$3^{2x}-3=0$ $3^{2x}+2= 0$ $x+3=0$

$3^{2x}=3$ ∅ $x=-3$

$2x=1$

$x=0.5$

+ - +
-------------(-3)----------------[0.5]-----------------
//////////////////////

$x$ ∈ $(-3;0.5]$