I need Math help please

Solve this system of equations:

(x-2)^2 + (y+1)^2 = 10
y=x-1

Substitute y=x-1 into (x-2)^2 + (y+1)^2 = 10
to get (x-2)^2 + (x-1+1)^2 = 10
(x-2)^2 + (x)^2 = 10
Multiply out to get
x^2 - 4x + 4 + x^2 = 10
Simplify
2x^2 - 4x - 6 = 0
divide by 2
x^2 - 2x - 3 = 0
factorise
(x +1)(x-3)=0
so x = -1 or x = 3
input the values of x into y=x-1 to get the values for y
y=-1-1=-2 so p1(-1,-2)
and
y=3-1 = 2 so p2(3,2)
To ensure our answers are correct it's always a good idea to input the values we have back into the formula to see if it makes sense.
-2 = -1 -1 (this one makes sense)
2 = 3 -1 (so does this so we are done.)

(x-2)^2 + (y+1)^2 = 10
y=x-1

you replace all the y in the formula by x-1

you would get: (x-2)^2 + (x-1+1)^2 = 10

now you develop the equation and you get

x^2-4x+4 + x^2 = 10

this is equal to

2x^2 -4x+4 =10
2x^2 -4x-6 = 0

Calculate the discriminant. (b^2-4*a*c)

(-4)^2 - 4(2)(-6) = 16 + 48 = 64

X1= (4+8)/4 = 3
x2 = (4-8)/4 = -1

x= 3 or -1

that means the first equation has a center of (2,-1) and a radius of square root of 10. Draw a circle (on graph paper) with a approximate radius of square root of ten and the center has to be (2,-1). then use any x variable and subtract 1 to get y. Draw that line. The solution of the system is the point(s) of intersection of the circle and line. Good Luck!