How find the equation of the tangent to the curve at a specific point? PLEASE HELP, 10 POINTS

A curve has equation x^2 + 2y^2 + 5x + 6y = 10. Find the equation of the tangent to the curve at the point (2, -1). Give your answer in the form ax+by+c=0, where a, b, and c are integers.
Please help me solve this, 10 points??

x^2 + 2y^2 + 5x + 6y = 10........(i)
Differentiate
2x +4y(dy/dx) +5+6(dy/dx) =0
(dy/dx){4y+6} =-2x-5
dy/dx = -{2x+5)/(4y+6)
Put x =2 and y=-1
dy/dx =-(4+5)/(-4+6) =-9/2.......(ii)
Therefore equation of tangent is
y+1 =-9/2(x-2)
2y+2 =-9x+18
9x+2y-16=0 .............Ans

Differentiate implicitly : 2x + 4yy' + 5 + 6y' = 0. Now plug in the given point and solve for y'.
4 - 4y' + 5 + 6y' = 0
y' = -9/2, the slope of the tangent line.
y = -9/2x + b
-1 = -9 + b; b = 8
9x + 2y - 16 = 0

y - x + 3 = 0?