Solve linear-quadratic system by substitution (circle, line)

[asissa]

Here are the two equations: x^2 + y^2 = 13 and 4y - 2x = 16. I have changed the second equation to: y = 1/2 x +4.
Using substitution, x^2 + (1/2x + 4)^2 = 13
x^2 + 1/4x^2 + 4x + 16 -13 = 0
5/4x^2 + 4x +3 = 0
x^2 + 16/5x + 12/5 = 0

And then I get lost. I know the answers, (-6/5, 17/5) and (-2, 3), but I don't understand where I am going wrong. Where have I made the mistake, or if I am right so far, where do I go now? No use of calculators, I can only use it to check.

I tried switching my equations around, hoping to get the (-2,3), but again, I am nowhere near it. Help please! Thank you.

 

[akoaysigod]

Re: Solve linear-quadratic system by substitution (circle, l

5/4x^2 + 4x +3 = 0
------->x^2 + 16/5x + 12/5 = 0

I think that last line is wrong. Instead of doing that, multiply the first one by 4 and you'll get

5x^2 + 16x + 12 this factors much more easily into (5x+6)(x+2) = 0 and you should be able to take it from there.