Identify second degree equation if the solutions are coordinates of the center of circle x^2+y^2+6y-4x=36.

We know that we can form the equation of second degree
when it's roots are known.

The quadratic is represented by
the general form:

x^2 - Sx + P = 0, where S = x1  + x2 (sum
of the roots) and P = x1*x2, product of roots.

Since the
roots of the quadratic are represented by the coordiates of the center of the given
circle, we need to determine them, first.

We'll create the
standad equation of the circle:

(x - h)^2 + (y - k)^2 =
r^2, where h and k are x and y coordinates of the center of circle and r is the
radius.

We'll take the terms in
x:

x^2 -4x

We'll add and
subtract 4:

x^2 -4x+4 - 4

The
first 3 terms represent the perfect square (x-2)^2

(x-2)^2
- 4

We'll take the terms in
y:

y^2 + 6y

We'll add and
subtract 9:

y^2 + 6y + 9 -
9

The first 3 terms represent the perfect square
(x+3)^2

(x+3)^2 - 9

We'll
re-write the equation of the circle:

(x-2)^2 - 4  + (x+3)^2
- 9 - 36 = 0

We'll combine like
terms:

(x-2)^2 + (x+3)^2 - 49 =
0

(x-2)^2 + (x+3)^2 =
49

Comparing, we'll get the coordinates of the center C (2
; -3) and the radius of the circle: r = 7

Now, since we
know the coordinates, we can create the quadratic
equation.

First, we'll determine the sum and the product of
the roots.

S = 2 - 3 = -1

P =
2*(-3) = -6

The quadratic equation, whose
roots are the coordinates of the center of given circle, is: x^2 + x - 6 =
0.