document.write( "Question 688050: How do I put the equation in standard form? The vertices and the center of the equation are also needed:\r
\n" ); document.write( "\n" ); document.write( "-16x^2-4y^2=48x-20y+57\r
\n" ); document.write( "\n" ); document.write( "I did try to solve it using completing the square but it didn't work out the correct way, because the numbers I came up with I was not able to factor them. \n" ); document.write( "
Algebra.Com's Answer #425490 by stanbon(75887)\"\" \"About 
You can put this solution on YOUR website!
put the equation in standard form? The vertices and the center of the equation are also needed:
\n" ); document.write( "-16x^2-4y^2 = 48x-20y+57
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\n" ); document.write( "Rearrange:
\n" ); document.write( "16x^2+48x + 4y^2 -20y + 57 = 0
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\n" ); document.write( "16(x^2+3x) + 4(y^2-5y) + 57 = 0
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\n" ); document.write( "Complete the square:
\n" ); document.write( "16(x^2 + 3x + (3/2)^2) + 4(y^2 - 5y + (5/2)^2) = -57 + 16(3/2)^2 + 4(5/2)^2
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\n" ); document.write( "16(x+(3/2))^2 + 4(y+(5/2))^2 = -57 + 36 + 25
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\n" ); document.write( "16(x+(3/2))^2/(56/16) + 4(y+(5/2)^2/(56/4) = 4
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\n" ); document.write( "(x+(3/2))^2/(1/4) + (y+(5/2))^2/1 = 1
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\n" ); document.write( "Center at (-3/2 , -5/2)
\n" ); document.write( "Vertices at (-3/2, (-5/2)-1) and (-3/2, (-5/2)+1)
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\n" ); document.write( "Cheers,
\n" ); document.write( "Stan H.
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