7.096. $\lg \left(3 x^{2}+12 x+19\right)-\lg (3 x+4)=1$.
## Solution.

Domain of definition: $3 x+4>0, x>-\frac{4}{3}$.

We have $\lg \frac{3 x^{2}+12 x+19}{3 x+4}=1, \frac{3 x^{2}+12 x+19}{3 x+4}=10,3 x^{2}-18 x-21=0$

for $3 x+4 \neq 0$. From this, $x_{1}=-1, x_{2}=7$.

Answer: $-1 ; 7$.