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Question

#9x^2 + 16y^2+36x-32y-92 = 0#

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Answer 1 · 2018-06-03T12:39:34

#x approx {-5.771, 1.771}#

#9x^2 + 16y^2+36x-32y-92 = 0#

#9x^2 + 16y^2+36x-32y = 92#

The above constitutes an ellipse as shown below.

graph{9x^2 + 16y^2+36x-32y = 92 [-10, 10, -5, 5]}

From which we can determine the #x-#intercepts as:

#x approx {-5.771, 1.771}#