You do not have to solve the simultaneous equation, but you need it.

From the data, we know that y = 4x -26.

This means the gradient (which is constant for a straight line) is 4 at any point (including point P)

We also know that we can obtain the gradient at point P by differentiating it, then substituting the x value of the point in. Thus

y' = 6x² - 6x -12 = 4 for point P (x,y)

Solve for x to find the horizontal value for P

You will get two different values of x

Now substitute that value of x into your simultaneous equation, and only one of them will satisfy the equation. That is the x value of point P.