Whoever can get me the correct answers to this will make me verrry happy :) 10 points for you too!

Find all points (both coordinates) on the given curve where the tangent line is horizontal. Round answers to three decimal places where necessary

19x^2 + xy + 3y = 2

x= and x=

So there are two answers, both x's with up to three decimal points...umm how on earth do you do this? I know you start with the dy/dx thing thats it though, then idk what you substitute to get the two answers for x

Then its fill in the blank. Since there ______ real solutions (are, are no), so there _____ (are, are no) horizontal tangents,

Good approach by nyc_kid.
Minor mistake in the quadratic equation, s/b
19x^2+114x+2=0
so,
x = (-114 +/- sqrt(12996-152))/38
x = (-114 +/- (113.331))/38
= -3 +/- 2.982

x_+ = -0.0176
x_- = -5.982


y_+ =-38*( -0.0176) = 0.669
y_- = -38*(-5.982) = 227.316

horizontal at
1) (-0.0176, 0.669)
2) (-5.982, 227.316)


Since there are 2 real solutions, so there are 2 horizontal tangents.

The implicit differentiation of the curve yields:
38x + y + xy' + 3y' = 0
You are looking for points with y' = 0 (horizontal tangents)
38x + y = 0 ==> y = -38x
Now go back to the original equation : 19x^2 + xy + 3y = 2, and plug in y = -38x:
19x^2 -38x^2 + 114x -2 = 0
19x^2 -114x + 2 = 0
Looks like QE, doesn't it?
x = [114 +/- sqrt(12996 - 152)] / 38
x = 5.982 y= -227.331
x = .0176 y = -.668