Could someone me with number 7 and 8? i might you and god bless you!

I don't see #8, but I can help with #7:

The student correctly set the problem up for using the quadratic formula. She got all all the terms on one side and zero on the other. She also ordered the problem in quadratic form, ax^2 + bx + c

She also correctly applied the quadratic formula, setting it up correctly based on the a, b, and c coefficients. 

Her mistake was between the third and fourth line: 
-4 (2) (-3) = 24, not -24, since there are two negative terms being multiplied. 
Therefore, it should be √(16 + 24) = √40, not √-8

This is a Systems of Equations problem, I'm guessing. 

6) To find the point of intersection, I have to first set the two equations together:


I can factor one side:


I can now solve for x to get the x-coordinate on the point of intersection (I just divided both sides by -1):


Solve for x:



Since it's a quadratic, I'll just use the quadratic equation to find the two x's. Remember, it says "find all break points" so there's more than 1:


and


Plug in:

and 


Solve both:

and


Now that we have the x-coordinates, we can plug both of them in into the equation that we had for y to find the point in form (x, y). So, let's do that, and since we can choose either equation, I'm going to choose the first one because it's much easier. Let's do it:


and



So, your two (and all of them) points of intersection are (7, 46) and (1, 14).
If any step doesn't make sense, hmu. I'm not going to do the other one because I don't have enough time. Maybe I can help you tomorrow in the DMs. 

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