Решите пожалуйста

$lo g _{7}(3-x)+lo g _{7}(4-10x)=2lo g _{7} 6$
ОДЗ:
$\left \{ {{3-x\ \textgreater \ 0} \atop {4-10x\ \textgreater \ 0}} \right.$
$\left \{ {{-x\ \textgreater \ -3} \atop {-10x\ \textgreater \ -4}} \right.$
$\left \{ {{x\ \textless \ 3} \atop {x\ \textless \ 0.4}} \right.$
$x$ ∈ $(-$ ∞ $;0.4)$

$lo g _{7}[(3-x)(4-10x)]=lo g _{7} 6^2$
$lo g _{7}(10x^2-34x+12)=lo g _{7} 36$
$10x^2-34x+12= 36$
$10x^2-34x+12- 36=0$
$10x^2-34x- 24=0$
$5x^2-17x- 12=0$
$D=(-17)^2-4*5*(-12)=289+240=529$
$x_1= \frac{17+23}{10}=4$ - ∅
$x_2= \frac{17-23}{10}=-0.6$

Ответ: -0,6