\(\Leftrightarrow2x^2-5x-12+x^2-7x+10=3x^2-17x+20\\ \Leftrightarrow5x=22\\ \Leftrightarrow x=\dfrac{22}{5}\)
\(\left(2x+3\right)\left(x-4\right)+\left(x-5\right)\left(x-2\right)=\left(3x-5\right)\left(x-4\right)\)
\(2x^2-8x+3x-12+x^2-2x-5x+10-3x^2+12x+5x-20=0\)
\(5x-22=0\)
\(x=\dfrac{22}{5}\)