Question 389516
To find the properties of the Conic solution:

4x^(2)-5y^(2)-16x-30y-9=0

Move all terms not containing a variable to the right-hand side of the equation.
4x^(2)-16x-5y^(2)-30y=9

To create a trinomial square on the left-hand side of the equation, add a value to both sides of the equation that is equal to the square of half the coefficient of x.  In this problem, add (-2)^(2) to both sides of the equation.
4(x^(2)-4x+4)-5y^(2)-30y=9+0+4*4

Factor the perfect trinomial square into (x-2)^(2).
4((x-2)^(2))-5y^(2)-30y=9+0+4*4

Factor the perfect trinomial square into (x-2)^(2).
4(x-2)^(2)-5y^(2)-30y=9+0+4*4

To create a trinomial square on the left-hand side of the equation, add a value to both sides of the equation that is equal to the square of half the coefficient of y.  In this problem, add (3)^(2) to both sides of the equation.
4(x-2)^(2)-5(y^(2)+6y+9)=9+0+4*4+0+9*-5

Factor the perfect trinomial square into (y+3)^(2).
4(x-2)^(2)-5((y+3)^(2))=9+0+4*4+0+9*-5

Factor the perfect trinomial square into (y+3)^(2).
4(x-2)^(2)-5(y+3)^(2)=9+0+4*4+0+9*-5

Multiply 4 by 4 to get 16.
4(x-2)^(2)-5(y+3)^(2)=9+0+16+0+9*-5

Multiply 9 by -5 to get -45.
4(x-2)^(2)-5(y+3)^(2)=9+0+16+0-45

Add 0 to 9 to get 9.
4(x-2)^(2)-5(y+3)^(2)=9+16+0-45

Add 16 to 9 to get 25.
4(x-2)^(2)-5(y+3)^(2)=25+0-45

Add 0 to 25 to get 25.
4(x-2)^(2)-5(y+3)^(2)=25-45

Subtract 45 from 25 to get -20.
4(x-2)^(2)-5(y+3)^(2)=-20

Multiply each term in the equation by -1.
(4(x-2)^(2)-5(y+3)^(2))*-1=-20*-1

Multiply (4(x-2)^(2)-5(y+3)^(2)) by -1 to get -(4(x-2)^(2)-5(y+3)^(2)).
-(4(x-2)^(2)-5(y+3)^(2))=-20*-1

Multiply -20 by -1 to get 20.
-(4(x-2)^(2)-5(y+3)^(2))=20

Divide each term by 20 to make the right-hand side equal to 1.
-(4(x-2)^(2)-5(y+3)^(2))/(20)=(20)/(20)

Simplify each term in the equation in order to set the right-hand side equal to 1.  The standard form of an ellipse or hyperbola requires the right-hand side of the equation be 1.
-(4(x-2)^(2)-5(y+3)^(2))/(20)=1


Let me know if this helps; if not, repost.