Lg(2-3x)+lg(2+3x)=lg(4-x)+lg(x) Помогите!

$lg(2-3x)+lg(2+3x)=lg(4-x)+lgx$
ОДЗ:
$\left \{ {{2-3x\ \textgreater \ 0} \atop {2+3x\ \textgreater \ 0}}}} \right.$
$\left \{ {{4-x\ \textgreater \ 0} \atop {x\ \textgreater \ 0}} \right.$

$\left \{ {{x\ \textless \ \frac{2}{3} } \atop {x\ \textgreater \ - \frac{2}{3} }}}} \right.$
$\left \{ {{x\ \textless \ 4} \atop {x\ \textgreater \ 0}} \right.$
$x$ $(0; \frac{2}{3} )$

$lg[(2-3x)(2+3x)]=lg[(4-x)x]$
$lg(4-9x^2)=lg(4x-x^2)$
$4-9x^2=4x-x^2$
$8x^2+4x-4=0$
$2x^2+x-1=0$
$D=1^2-4*2*(-1)=9$
$x_1= \frac{-1+3}{4}=0.5$
$x_2= \frac{-1-3}{4}=-1$ ∅

Ответ: 0.5