\(5+4x-x+2=\left(5x+4\right)\left(7+5x\right)\)
\(\Leftrightarrow5+4x-x+2=35+28x+25x+20x^2\)
\(\Leftrightarrow x^2+50x+28=0\)
Ta có \(\Delta=50^2-4.1.28=2388,\sqrt{\Delta}=2\sqrt{597}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{-50+2\sqrt{597}}{2}=-25+\sqrt{597}\\x=\frac{-50-2\sqrt{597}}{2}=-25-\sqrt{597}\end{cases}}\)
\(5+4x-x+2=\left(5+4x\right)\left(7+5x\right)\)
\(7+3x=\left(5+4x\right)\left(7+5x\right)\)
\(7+3x=35+28x+25x+20x^2\)
\(7+3x-35-28x-25x-20x^2=0\)
\(-28-50x-20x^2=0\)
\(-28-50x-20x^2=0\)
\(x=-\frac{25+\sqrt{65}}{20};-\frac{25-\sqrt{65}}{20}\)
