Question

Complete the square
-x^2-y^2-3x+2y+4=0

Answer 1

\[-x^2 - y^2 - 3x + 2y + 4 = 0\]\[x^2 + y^2 + 3x - 2y - 4 = 0\]\[x^2 + y^2 + 3x - 2y = 4\]\[x^2 + 3x + \frac{9}{4} + y^2 - 2y + 1 = 4 + \frac{9}{4} + 1\]\[\left( x + \frac{3}{2} \right)^2 + \left( y - 1 \right)^2 = \frac{29}{4}\]

Answer 2

Thank you once again!!!