Question 145040
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^2-4y^2+2x-32y-67=0}}} 
<pre><font size = 4 color = "indigo"><b>
Rearrange so that the {{{y^2}}} term is first,
then the {{{y}}} term, then the others:

{{{-4y^2-32y+x^2+2x-67=0}}}

Group the last three terms on the left:

{{{-4y^2-32y+(x^2+2x-67)=0}}}

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

{{{4y^2+32y-(x^2+2x-67)=0}}}

Now we will use the quadratic formula with

{{{A = 4}}}, {{{B=32}}}, {{{C = -(x^2+2x-67)}}}

{{{y = (-B +- sqrt( B^2-4AC ))/(2A) }}}

{{{y = (-32 +- sqrt( 32^2-4(4)(-(x^2+2x-67)) ))/(2(4)) }}}

{{{y = (-32 +- sqrt(1024-16(-x^2-2x+67) ))/8 }}}

{{{y = (-32 +- sqrt(1024+16x^2+32x-1072) )/8 }}}

{{{y = (-32 +- sqrt(16x^2+32x-48) )/8 }}}

{{{y = (-32 +- sqrt(16(x^2+2x-3)) )/8 }}}

{{{y = (-32 +- sqrt(16)sqrt(x^2+2x-3)) /8 }}}

{{{y = (-32 +- 4sqrt(x^2+2x-3)) /8 }}}

{{{y = (4(-8 +- sqrt(x^2+2x-3))) /8 }}}

{{{y = (-8 +- sqrt(x^2+2x-3)) /2 }}}

Edwin</pre>