(x+2)²+(y-3)=0

To solve the equation, we need to simplify and rearrange it to find the values of x and y that satisfy the equation. Starting with the given equation:

(x+2)²+(y-3)=0

First, let's expand the square of (x+2):

(x+2)² = (x+2)(x+2) = x² + 4x + 4

Now, let's rewrite the equation with the expanded square:

x² + 4x + 4 + (y-3) = 0

Next, let's simplify the equation:

x² + 4x + y + 1 = 0

Now, to solve for x, we can use the quadratic formula:

x = (-b ± √(b² - 4ac)) / (2a)

Comparing the equation to ax² + bx + c = 0, we can identify the values of a, b, and c as:

a = 1, b = 4, and c = (y + 1)

Substituting these values into the quadratic formula, we get:

x = (-4 ± √(4² - 4(1)(y + 1))) / (2(1)) x = (-4 ± √(16 - 4(y + 1))) / 2 x = (-4 ± √(16 - 4y - 4)) / 2 x = (-4 ± √(12 - 4y)) / 2 x = (-4 ± 2√(3 - y)) / 2 x = -2 ± √(3 - y)

Therefore, the solution for x is -2 ± √(3 - y).

Now, let's solve for y:

(x+2)² + (y-3) = 0 (x+2)² = -y + 3 (x+2)² - 3 = -y 3 - (x+2)² = y

Therefore, the solution for y is y = 3 - (x+2)².

In summary, the solutions to the equation (x+2)² + (y-3) = 0 are x = -2 ± √(3 - y) and y = 3 - (x+2)².