Two parabolas are the graphs of the equations $y=2x^2-10x-10$ and $y=x^2-4x+6$. give all points where they intersect. list the points in order of increasing $x$-coordinate, separated by semicolons.

(-2, 18) and (8, 38)

Step-by-step explanation:

First, set the two equations equal to each other to get $2x^2-10x-10=x^2-4x+6$. Combine like terms to get $x^2-6x=16$. To complete the square, we need to add $\left(\dfrac{6}{2}\right)^2=9$ to both sides, giving $(x-3)^2=16+9=25$.

So we have $x-3=\pm5$. Solving for $x$ gives us $x=-2$ or $8$. Using these in our original parabolas, we find the points of intersection to be $\boxed{(-2,18)}$ and $\boxed{(8,38)}$.

Credit: AoPs

answer: the answer is 127 degrees

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step-by-step explanation:

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