Set the function equal to zero and clear the denominator by multiplying by the least common denominator, \(\left( {w + 1} \right)\left( {2w - 3} \right)\), and then solve the resulting equation.

\[\begin{align*}\left( {w + 1} \right)\left( {2w - 3} \right)\left( {\frac{{2w}}{{w + 1}} + \frac{{w - 4}}{{2w - 3}}} \right) & = 0\\ 2w\left( {2w - 3} \right) + \left( {w - 4} \right)\left( {w + 1} \right) & = 0\\ 5{w^2} - 9w - 4 & = 0\end{align*}\]

This quadratic doesn’t factor so we’ll need to use the quadratic formula to get the solution.

\[w = \frac{{9 \pm \sqrt {{{\left( { - 9} \right)}^2} - 4\left( 5 \right)\left( { - 4} \right)} }}{{2\left( 5 \right)}} = \frac{{9 \pm \sqrt {161} }}{{10}}\]

So, it looks like this function has the following two roots,

\[\frac{{9 + \sqrt {161} }}{{10}} = 2.168858\hspace{0.25in}\hspace{0.25in}\hspace{0.25in}\frac{{9 - \sqrt {161} }}{{10}} = - 0.368858\]

Recall that because we started off with a function that contained rational expressions we need to go back to the original function and make sure that neither of these will create a division by zero problem in the original function. Neither of these do and so they are the two roots of this function.