Algebra 2
posted by Soto on .
16x^2+16y^216x+24y3=0
How do I solve this?

"Solving" means finding ordered pairs that satisfy the equation.
Your equation is a circle, and there are an infinite number of ordered pairs that will work.
Did you mean to put it in the standard form of a circle equation? 
yes

16x^2+16y^216x+24y3=0
16(x^2  x + .....) + 16(y^2 + (3/2)y + ... = 3
16(x^2  x + 1/4) + 16(y^2 + (3/2)y + 9/16) = 3 + 4 + 9
16(x1/2)^2 + 16(y+3/4) = 16
(x1/2)^2 + (y+3/4)^2 = 1
all vital info about the circle is now obvious. 
thank you very much

The coordinate of all relative maxima 2x^4+x^333x^216x+16