Sum of all real roots of $\frac{{3{x^2} - 9x + 17}}{{{x^2} + 3x + 10}} = \frac{{5{x^2} - 7x + 19}}{{3{x^2} + 5x + 12}}$? 
The sum of real roots of $\dfrac{{3{x^2} - 9x + 17}}{{{x^2} + 3x + 10}} = \dfrac{{5{x^2} - 7x + 19}}{{3{x^2} + 5x + 12}}$ is____


How do I proceed with this type of problem.
Cross multiplying and segregating the coefficient is a cumbersome process
My putting in desmos.com my answer is $«6»$.

Cross-multiplying gives:
$$(3x^2 - 9x + 17)(3x^2 + 5x + 12) = (x^2 + 3x + 10)(5x^2 - 7x + 19)$$
$$9x^4 + 15x^3 + 36x^2 - 27x^3 - 45x^2 - 108x + 51x^2 + 85x + 204 = 5x^4 - 7x^3 + 19x^2 + 15x^3 - 21x^2 + 57x + 50x^2 - 70x + 190$$
$$9x^4 - 12x^3 + 42x^2 - 23x + 204 = 5x^4 + 8x^3 + 48x^2 - 13x + 190$$
$$4x^4 - 20x^3 - 6x^2 - 10x + 14 = 0$$
$$x^4 - 5x^3 - \frac{3}{2}x^2 - \frac{5}{2}x + \frac{7}{2} = 0$$
There, that wasn't too bad.  Now the hard part: finding the roots.
The Rational Root Theorem tells us that all rational roots are in $\pm\lbrace 1, 7, \frac{1}{2}, \frac{7}{2} \rbrace$.  Unfortunately, it turns out that none of those actually are roots, so let's take another path and try a quadratic factorization:
$$f(x) = (x^2 + Ax + B)\left(x^2 - (A + 5)x + \frac{7}{2B}\right)$$
where the coefficients in the second factor are set to make the cubic ($-5$) and constant ($\frac{7}{2}$) coefficients correct.  Expanding the product:
$$f(x) = x^4 - (A + 5)x^3 + \frac{7}{2B}x^2 + Ax^3 - A(A + 5)x^2 + \frac{7A}{2B}x + Bx^2 - (A + 5)Bx + \frac{7}{2}$$
$$f(x) = x^4 - 5x^3 + \left(\frac{7}{2B} - A^2 - 5A + B \right)x^2 + \left(\frac{7A}{2B} - AB - 5B\right)x + \frac{7}{2}$$
Solving for the $x^2$ and $x$ coefficients is a bit difficult, but it turns out that $A = B = 1$ works, so:
$$f(x) = (x^2 + x + 1)\left(x^2 - 6x + \frac{7}{2}\right) = 0$$
The first factor has a negative discriminant (-3) and thus no real solutions, but the second one has a positive discriminant (22).
We don't actually have to find the roots, just use the fact that the two roots of a quadratic $ax^2 + bx + c = 0$ add up to $\frac{-b}{a}$.
So the answer to your question is indeed 6.
