OK, now that we've gotten that out of the way...
Let [itex](x_0, y_0)[/itex] be the point of tangency on the graph of the curve. BTW, you have drawn the graph, right?
At the point of tangency, the tangent line has to extend from [itex](x_0, y_0)[/itex] to (3, 14).
Here is an outline of the steps you'll need to carry out for this problem:
1. Find the slope of the line from [itex](x_0, y_0) = (x_0, x_0^2 + 2x_0)[/itex] to (3, 14).
2. By calculating the derivative and evaluating it at [itex]x_0[/itex], find the slope of the tangent line.
3. Equate the value you got in step 1 with the value from step 2, and solve for [itex]x_0[/itex]. (I got two values for [itex]x_0[/itex].)
4. Find the associated y value for each value of [itex]x_0[/itex] from step 3.
5. Using each point [itex](x_0, y_0)[/itex], find the equation of the line from [itex](x_0, y_0)[/itex] to (3, 14). There are two distinct equations.
Is that enough of a hint?