SOLUTION: {{{x^2-4y^2+2x-32y-67=0}}} I need this equation in y form pleas and thank u and please show all the steps of how to get y by itself.

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: {{{x^2-4y^2+2x-32y-67=0}}} I need this equation in y form pleas and thank u and please show all the steps of how to get y by itself.       Log On


   

Question 145040: x%5E2-4y%5E2%2B2x-32y-67=0
I need this equation in y form pleas and thank u and please show all the steps of how to get y by itself.
Answer by Edwin McCravy(20060) About Me  (Show Source):
You can put this solution on YOUR website!
I need this equation in y form please and thank you
and please show all the steps of how to get y by itself.
x%5E2-4y%5E2%2B2x-32y-67=0 

Rearrange so that the y%5E2 term is first,
then the y term, then the others:

-4y%5E2-32y%2Bx%5E2%2B2x-67=0

Group the last three terms on the left:

-4y%5E2-32y%2B%28x%5E2%2B2x-67%29=0

It will be easier if we multiply every term
through by -1, so it will begin with a
positive term:

4y%5E2%2B32y-%28x%5E2%2B2x-67%29=0

Now we will use the quadratic formula with

A+=+4, B=32, C+=+-%28x%5E2%2B2x-67%29

y+=+%28-B+%2B-+sqrt%28+B%5E2-4AC+%29%29%2F%282A%29+

y+=+%28-32+%2B-+sqrt%28+32%5E2-4%284%29%28-%28x%5E2%2B2x-67%29%29+%29%29%2F%282%284%29%29+

y+=+%28-32+%2B-+sqrt%281024-16%28-x%5E2-2x%2B67%29+%29%29%2F8+

y+=+%28-32+%2B-+sqrt%281024%2B16x%5E2%2B32x-1072%29+%29%2F8+

y+=+%28-32+%2B-+sqrt%2816x%5E2%2B32x-48%29+%29%2F8+

y+=+%28-32+%2B-+sqrt%2816%28x%5E2%2B2x-3%29%29+%29%2F8+

y+=+%28-32+%2B-+sqrt%2816%29sqrt%28x%5E2%2B2x-3%29%29+%2F8+

y+=+%28-32+%2B-+4sqrt%28x%5E2%2B2x-3%29%29+%2F8+

y+=+%284%28-8+%2B-+sqrt%28x%5E2%2B2x-3%29%29%29+%2F8+

y+=+%28-8+%2B-+sqrt%28x%5E2%2B2x-3%29%29+%2F2+

Edwin