Write the equation -4x^2+9y^2+32x+36y-64=0 in standard form. Please show me each step of the process! Thank you!

Hey there, hope I can help!

[tex]-4x^2+9y^2+32x+36y-64=0[/tex]

[tex]\mathrm{Add\:}64\mathrm{\:to\:both\:sides} \ \textgreater \ 9y^2+32x+36y-4x^2=64[/tex]

[tex]\mathrm{Factor\:out\:coefficient\:of\:square\:terms} \ \textgreater \ -4\left(x^2-8x\right)+9\left(y^2+4y\right)=64[/tex]

[tex]\mathrm{Divide\:by\:coefficient\:of\:square\:terms:\:}4 [/tex]
[tex]-\left(x^2-8x\right)+\frac{9}{4}\left(y^2+4y\right)=16[/tex]

[tex]\mathrm{Divide\:by\:coefficient\:of\:square\:terms:\:}9 [/tex]
[tex]-\frac{1}{9}\left(x^2-8x\right)+\frac{1}{4}\left(y^2+4y\right)=\frac{16}{9}[/tex]

[tex]\mathrm{Convert}\:x\:\mathrm{to\:square\:form} [/tex]
[tex]-\frac{1}{9}\left(x^2-8x+16\right)+\frac{1}{4}\left(y^2+4y\right)=\frac{16}{9}-\frac{1}{9}\left(16\right)[/tex]

[tex]\mathrm{Convert\:to\:square\:form}[/tex]
[tex]-\frac{1}{9}\left(x-4\right)^2+\frac{1}{4}\left(y^2+4y\right)=\frac{16}{9}-\frac{1}{9}\left(16\right)[/tex]

[tex]\mathrm{Convert}\:y\:\mathrm{to\:square\:form}[/tex]
[tex]-\frac{1}{9}\left(x-4\right)^2+\frac{1}{4}\left(y^2+4y+4\right)=\frac{16}{9}-\frac{1}{9}\left(16\right)+\frac{1}{4}\left(4\right)[/tex]

[tex]\mathrm{Convert\:to\:square\:form}[/tex]
[tex]-\frac{1}{9}\left(x-4\right)^2+\frac{1}{4}\left(y+2\right)^2=\frac{16}{9}-\frac{1}{9}\left(16\right)+\frac{1}{4}\left(4\right)[/tex]

[tex]\mathrm{Refine\:}\frac{16}{9}-\frac{1}{9}\left(16\right)+\frac{1}{4}\left(4\right) \ \textgreater \ -\frac{1}{9}\left(x-4\right)^2+\frac{1}{4}\left(y+2\right)^2=1[/tex]

[tex]Refine\;once\;more\;-\frac{\left(x-4\right)^2}{9}+\frac{\left(y+2\right)^2}{4}=1[/tex]

For me I used
[tex]\frac{\left(y-k\right)^2}{a^2}-\frac{\left(x-h\right)^2}{b^2}= 1[/tex]
[tex]As\;\mathrm{it\;\:is\:the\:standard\:equation\:for\:an\:up-down\:facing\:hyperbola}[/tex]

I know yours is an equation which is why I did not go any further because this is the standard form you are looking for. I would rewrite mine to get my hyperbola standard form. However the one I have provided is the form you need where mine would be.
[tex]\frac{\left(y-\left(-2\right)\right)^2}{2^2}-\frac{\left(x-4\right)^2}{3^2}=1[/tex]

Hope this helps!