\(\left(x+4\right)\left(3x+4\right)-3x^2=176\)
\(\Leftrightarrow\text{}\text{}\text{}\text{}\text{}\text{}3x^2+4x+12x+16-3x^2=176\)
\(\Leftrightarrow\left(3x^2-3x^2\right)+\left(12x+4x\right)=176-16\)
\(\Leftrightarrow16x=160\)
\(\Leftrightarrow x=\dfrac{160}{16}\)
\(\Leftrightarrow x=10\)