### Instruction: Two parabolas are the graphs of the equations $y=2x^2-10x-10$ and $y=x^2-4x+6$. Find all points where they intersect. List the points in order of increasing $x$-coordinate, separated by semicolons.### Response: To find the points of intersection, I need to solve the system of equations $y=2x^2-10x-10$ and $y=x^2-4x+6$.
I can do this by equating the two expressions for $y$ and simplifying.
So I have $2x^2-10x-10=x^2-4x+6$.
Subtracting $x^2-4x+6$ from both sides, I get $x^2-6x-16=0$.
I can factor this quadratic equation as $(x-8)(x+2)=0$.
So the solutions are $x=8$ and $x=-2$.
To find the corresponding $y$-values, I can plug these $x$-values into either equation.
I'll use the first one.
When $x=8$, I get $y=2(8)^2-10(8)-10=2(64)-80-10=128-90=38$.
When $x=-2$, I get $y=2(-2)^2-10(-2)-10=2(4)+20-10=8+10=18$.
So the points of intersection are $(-2,18)$ and $(8,38)$.