The sum of all real values of $$x$$ for which $$\frac{3x^2 - 9x + 17}{x^2 + 3x + 10} = \frac{5x^2 - 7x + 19}{3x^2 + 5x + 12}$$ is equal to

The given equation is:
$$\frac{3x^2 - 9x + 17}{x^2 + 3x + 10} = \frac{5x^2 - 7x + 19}{3x^2 + 5x + 12}$$
Cross multiplying both sides gives:
$$(3x^2 - 9x + 17)(3x^2 + 5x + 12) = (5x^2 - 7x + 19)(x^2 + 3x + 10)$$
Let:
$$A = 3x^2 - 9x + 17$$
$$B = 3x^2 + 5x + 12$$
$$C = x^2 + 3x + 10$$
Expressing $$5x^2 - 7x + 19$$ in terms of $$A$$, $$B$$, and $$C$$:
$$A + B - C = (3x^2 - 9x + 17) + (3x^2 + 5x + 12) - (x^2 + 3x + 10)$$
$$A + B - C = 5x^2 - 7x + 19$$
Substituting these variables into the cross multiplied equation:
$$AB = (A + B - C)C$$
Expanding and rearranging terms:
$$AB = AC + BC - C^2$$
$$AB - AC - BC + C^2 = 0$$
Factoring by grouping:
$$A(B - C) - C(B - C) = 0$$
$$(A - C)(B - C) = 0$$
Now evaluating $$A - C$$ and $$B - C$$:
First factor:
$$A - C = (3x^2 - 9x + 17) - (x^2 + 3x + 10) = 2x^2 - 12x + 7$$
Second factor:
$$B - C = (3x^2 + 5x + 12) - (x^2 + 3x + 10) = 2x^2 + 2x + 2$$
Thus, the factored equation becomes:
$$(2x^2 - 12x + 7)(2x^2 + 2x + 2) = 0$$
Case 1:
$$2x^2 + 2x + 2 = 0$$
$$x^2 + x + 1 = 0$$
Checking discriminant $$D$$:
$$D = 1^2 - 4(1)(1) = -3$$
Since $$D < 0$$, there are no real roots for this factor.
Case 2:
$$2x^2 - 12x + 7 = 0$$
Checking discriminant $$D$$:
$$D = (-12)^2 - 4(2)(7) = 144 - 56 = 88$$
Since $$D > 0$$, this quadratic equation yields two real roots.
Using Vieta's relation, the sum of roots for $$2x^2 - 12x + 7 = 0$$ is:
$$\text{Sum of roots} = -\frac{-12}{2} = 6$$
Hence, the sum of all real values of $$x$$ is 6.