Giới hạn đã cho hữu hạn nên \(ax^2+\left(2b-3\right)x+5=0\) có nghiệm \(x=2\)
\(\Rightarrow4a+2\left(2b-3\right)+5=0\Rightarrow4a+4b-1=0\)
\(\Rightarrow2b=\dfrac{1-4a}{2}\)
\(\Rightarrow\lim\limits_{x\rightarrow2}\dfrac{ax^2+\left(\dfrac{1-4a}{2}-3\right)x+5}{\left(x-2\right)\left(x+3\right)}=\lim\limits_{x\rightarrow2}\dfrac{2ax^2-4ax-5x+10}{2\left(x-2\right)\left(x+3\right)}=\lim\limits_{x\rightarrow2}\dfrac{\left(x-2\right)\left(2ax-5\right)}{2\left(x-2\right)\left(x+3\right)}\)
\(=\lim\limits_{x\rightarrow2}\dfrac{2ax-5}{2\left(x+3\right)}=\dfrac{4a-5}{10}=10\Rightarrow a=\dfrac{105}{4}\)
\(\Rightarrow b=\dfrac{1-4a}{4}=-26\Rightarrow a+2b=-\dfrac{103}{4}\)
