Two parabolas are the graphs of the equations $y=3x^2+4x-5$ and $y=x^2+11$. Give all points where they intersect. List the points in order of increasing $x$-coordinate, separated by semicolons.
Setting the right-hand sides of the given equations equal gives $3x^2+4x-5=x^2+11$. Combining like terms gives $2x^2+4x=16$. Dividing by $2$ gives $x^2+2x=8$, and rearranging gives $x^2 +2x - 8=0$.  Factoring gives $(x+4)(x-2)=0$, so our solutions are $x=-4$ and $x=2$.  Substituting these into either of the original equations to find the corresponding values of $y$, we find the points of intersection to be $\boxed{(-4, 27);(2, 15)}$.
(-4,27);(2,15)
Level 5
Algebra
Problem: Two parabolas are the graphs of the equations $y=3x^2+4x-5$ and $y=x^2+11$. Give all points where they intersect. List the points in order of increasing $x$-coordinate, separated by semicolons.
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the points of intersection of the two parabolas, set the equations equal to each other: 

3x^2 + 4x - 5 = x^2 + 11. 

Rearranging this equation leads to:

3x^2 - x^2 + 4x - 5 - 11 = 0, 

which simplifies to:

2x^2 + 4x - 16 = 0. 

Next, divide the entire equation by 2 to simplify:

x^2 + 2x - 8 = 0. 

Now, apply the quadratic formula, x = (-b ± √(b^2 - 4ac)) / 2a, where a = 1, b = 2, and c = -8. 

Calculate the discriminant:

b^2 - 4ac = 2^2 - 4(1)(-8) = 4 + 32 = 36. 

Since the discriminant is positive, there are two real solutions. 

Now, substitute into the quadratic formula:

x = (-2 ± √36) / 2(1) = (-2 ± 6) / 2. 

This results in two values for x:

x = (4) / 2 = 2 and x = (-8) / 2 = -4. 

Next, substitute these x-values back into either original equation to find the corresponding y-values. 

For x = 2:

y = 3(2)^2 + 4(2) - 5 = 3(4) + 8 - 5 = 12 + 8 - 5 = 15. 

For x = -4:

y = 3(-4)^2 + 4(-4) - 5 = 3(16) - 16 - 5 = 48 - 16 - 5 = 27. 

Thus, the points of intersection are (-4, 27) and (2, 15). 

List the points in order of increasing x-coordinate, resulting in the final answer: (-4, 27); (2, 15).<|END_OF_LATENT|> The final answer is $(-4,27);(2,15)$.