how do you solve this problem?

Question

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

Answer 1

x^2 -x -6 = x^2 + 6x + 9

-x-6 = 6x + 9

-7x = 15

x = -15/7

Answer 2

Open parentheses in both sides:

x^2 -x - 6 = x^2 + 6x + 9

-6 - 9 = 6x + x

-15 = 7x

x = -15/7

Regards
Tonio

Answer 3

I will tell you how. I will not solve it.
Multiply out every multiplication shown:
On the left, each parenthesized term by the other
On the right, the parenthesized term by itself.
Move the terms all to the left side of "=".

Put the left side in the form of ax^2 + bx + c = 0 so that you can identify what "a", "b", and "c" are in this problem.

Plug the values of "a", "b", and "c" into the proper places in the "quadratic (roots) formula, which should have been given by your teacher or textbook. Then you calculate the two roots from that formula.

Answer 4

(x+2)(x-3) = (x+3)^2
x^2 - x - 6 = x^2 + 6x + 9

to get from there, use FOIL as your aid.
FOIL stands for:
First
Outside
Inside
Last

x^2 - x - 6 = x^2 + 6x + 9
Now, subtract x^2 so that they'll cancel each other out
now you have -x-6=6x+9
get the x's on one side and the constants on the other
by adding 6 and subtracting 6x from both sides
-x-6x=6+9
combine like terms
-7x=15
divide by -7
x=-15/7

Answer 5

x² - x - 6 = x² + 6x + 9
- 15 = 7x
x = - 15 / 7

Answer 6

(x+2)(x-3) = (x+3)^2
x^2 + 2x - 3x - 6 = x^2 + 3x +3x + 9
-x - 6 = 6x + 9
-15 = 7x
x = -(15/7)

Answer 7

(x+2)(x-3)=(x+3)^2
x^2-x-6=x^2+6x+9
-7x=14
x=-2