Solve the expression;
$\dfrac{x+3}{x-2}-\dfrac{1-x}{x}=\dfrac{17}{x}\left( x\ne 0,2 \right)$

Hint: We solve this question by bringing the term in RHS to LHS. Then we simplify the terms with the same denominator. Then we take the LCM of the denominators of the two fractions and multiply each fraction with corresponding variables to get the denominator of both fractions as LCM. Then we add the numerators and then simplify them to get an equation. Then we find the roots of the equation using the formula for the roots of the quadratic equation \[a{{x}^{2}}+bx+c=0\], $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$. Then we simplify the obtained values to find the values of $x$.

Complete step-by-step solution
We are given that $\dfrac{x+3}{x-2}-\dfrac{1-x}{x}=\dfrac{17}{x}$.
We are also given that $x\ne 0,2$.
So, now let us consider the given equation.
$\Rightarrow \dfrac{x+3}{x-2}-\dfrac{1-x}{x}=\dfrac{17}{x}$
Now let us bring the expression $\dfrac{17}{x}$ to the LHS. Then the equation becomes,
\[\begin{align}
  & \Rightarrow \dfrac{x+3}{x-2}-\dfrac{1-x}{x}-\dfrac{17}{x}=0 \\
 & \Rightarrow \dfrac{x+3}{x-2}-\left( \dfrac{1-x}{x}+\dfrac{17}{x} \right)=0 \\
\end{align}\]
As the denominators of the fractions inside the bracket are the same, we can add the numerators. Then we get,
\[\begin{align}
  & \Rightarrow \dfrac{x+3}{x-2}-\left( \dfrac{1-x+17}{x} \right)=0 \\
 & \Rightarrow \dfrac{x+3}{x-2}-\dfrac{18-x}{x}=0 \\
\end{align}\]
Now taking the LCM of denominators we get LCM as $x\left( x-2 \right)$.
So, now we can multiply each fraction accordingly to make the denominator as LCM, we get,
\[\Rightarrow \dfrac{x\left( x+3 \right)}{x\left( x-2 \right)}-\dfrac{\left( 18-x \right)\left( x-2 \right)}{x\left( x-2 \right)}=0\]
Adding them we get,
\[\Rightarrow \dfrac{x\left( x+3 \right)-\left( 18-x \right)\left( x-2 \right)}{x\left( x-2 \right)}=0\]
As we are given that $x\ne 0,2$ we can write the above equation as,
\[\Rightarrow x\left( x+3 \right)-\left( 18-x \right)\left( x-2 \right)=0\]
Simplifying it we get,
\[\begin{align}
  & \Rightarrow {{x}^{2}}+3x-\left( 18x-36+2x-{{x}^{2}} \right)=0 \\
 & \Rightarrow {{x}^{2}}+3x-\left( 20x-36-{{x}^{2}} \right)=0 \\
\end{align}\]
Now we can write it as,
\[\begin{align}
  & \Rightarrow {{x}^{2}}+3x-20x+36+{{x}^{2}}=0 \\
 & \Rightarrow 2{{x}^{2}}-17x+36=0 \\
\end{align}\]
Now let us consider the formula for the roots of the quadratic equation \[a{{x}^{2}}+bx+c=0\].
$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
Now comparing the equation \[2{{x}^{2}}-17x+36=0\] with \[a{{x}^{2}}+bx+c=0\] we get that
\[\Rightarrow a=2,b=-17,c=36\]
Substituting these values in the above formula we get,
$\begin{align}
  & \Rightarrow x=\dfrac{-\left( -17 \right)\pm \sqrt{{{\left( -17 \right)}^{2}}-4\left( 2 \right)\left( 36 \right)}}{2\left( 2 \right)} \\
 & \Rightarrow x=\dfrac{17\pm \sqrt{289-288}}{4} \\
 & \Rightarrow x=\dfrac{17\pm \sqrt{1}}{4} \\
 & \Rightarrow x=\dfrac{17\pm 1}{4} \\
\end{align}$
So, we get the values of $x$ as,
$\begin{align}
  & \Rightarrow x=\dfrac{17+1}{4},x=\dfrac{17-1}{4} \\
 & \Rightarrow x=\dfrac{18}{4},x=\dfrac{16}{4} \\
 & \Rightarrow x=\dfrac{9}{2},x=4 \\
\end{align}$
So, we get that the values satisfying the given equation $\dfrac{x+3}{x-2}-\dfrac{1-x}{x}=\dfrac{17}{x}\left( x\ne 0,2 \right)$ are $\dfrac{9}{2},4$.
Hence the answer is $x=\dfrac{9}{2},4$.

Note: We can also solve the quadratic equation obtained above in the solution by factorization.
The equation obtained is,
\[\Rightarrow 2{{x}^{2}}-17x+36=0\]
We can write it as,
\[\begin{align}
  & \Rightarrow 2{{x}^{2}}-8x-9x+36=0 \\
 & \Rightarrow 2x\left( x-4 \right)-9\left( x-4 \right)=0 \\
 & \Rightarrow \left( 2x-9 \right)\left( x-4 \right)=0 \\
 & \Rightarrow x=\dfrac{9}{2},4 \\
\end{align}\]
Hence the answer is $x=\dfrac{9}{2},4$.