Hello, batticaloa!
Sorry, but I have to say this . . .
\(\displaystyle \;\;\)You're in Calculus, but you can't solve a quadratic?
\(\displaystyle \L y\;=\;\frac{-5x^2\,+\,3x}{2x^2\,-\,5}\)
\(\displaystyle \L y'\;=\;\frac{-6x^2\,+\,50x\,-\,15}{(2x\,-\,5)^2}\)
We have: \(\displaystyle \,-6x^2\,+\,50x\,-\,15\:=\:0\;\;\Rightarrow\;\;6x^2\,-\,50x\,+\,15\:=\:0\)
\(\displaystyle \;\;\;\;\;\;\;\;a\,=\,6,\;b\,=\,-50,\;c\,=\,15\)
Quadratic Formula: \(\displaystyle \;x\;=\;\frac{-(-50)\,\pm\,\sqrt{(-50)^2\,-\,4(6)(15)}}{2(6)} \;=\;\frac{50\,\pm\,\sqrt{2140}}{12}\;=\;\frac{50\,\pm\,2\sqrt{535}}{12}\)
And we have: \(\displaystyle \:\begin{array}{cc}x_1\:=\:\frac{25\,+\,\sqrt{535}}{6}\:=\:8.021677835 \\ x_2\:=\:\frac{25\,-\,\sqrt{535}}{6}\:=\:0.311655498\end{array}\;\;\) . . . critical values
\(\displaystyle \;\;\;y''\;=\;\frac{(2x\,-\,5)^2(-12x\.+\,50)\,-\,(-6x^2\,+\,50x\,-\,15)\cdot2(2x\,-\,5)\cdot 2}{(2x\,-\,5)^4}\)
Factor: \(\displaystyle \,(2x\,-\,5)\,\frac{(2x\,-\,5)(-12x\,+\,50)\,-\,4(-6x^2\,+\,50x\,-\,15)}{(2x\,-\,5)^4}\)
\(\displaystyle \;\;\;=\;\frac{-24x^2\,+\,100x\,+\,60x\,-\,250\,+\,24x^2\,-\,200x\,+\,60}{(2x\,-\,5)^3} \;= \;\frac{-40x\,-\,190}{(2x\,-\,5)^3}\)
Hence: \(\displaystyle \L\,y''\;=\;-10\,\frac{4x\,+\,19}{(2x\,-\,5)^3}\;\;\) . . . etc.
