(х+2)²+2(х+2)-3=0

To solve this equation, we can use the quadratic formula or complete the square.

Using the quadratic formula: The equation we have is in the form of ax^2 + bx + c = 0, where a = 1, b = 4, and c = -1.

The quadratic formula is given by: x = (-b ± √(b^2 - 4ac)) / (2a)

Plugging in the values, we get: x = (-(2) ± √((2)^2 - 4(1)(-3))) / (2(1)) x = (-2 ± √(4 + 12)) / 2 x = (-2 ± √16) / 2 x = (-2 ± 4) / 2

This gives us two possible solutions: x = (-2 + 4) / 2 = 2/2 = 1 x = (-2 - 4) / 2 = -6/2 = -3

Therefore, the solutions to the equation are x = 1 and x = -3.

Alternatively, we can complete the square:

Starting with the equation: (x+2)² + 2(x+2) - 3 = 0

Expanding the square: x² + 4x + 4 + 2x + 4 - 3 = 0 x² + 6x + 5 = 0

To complete the square, we need to add the square of half the coefficient of x to both sides: x² + 6x + 9 + 5 = 9 (x + 3)² + 5 = 9

Simplifying: (x + 3)² = 4

Taking the square root of both sides: x + 3 = ±√4 x + 3 = ±2

Solving for x: x = -3 + 2 = -1 x = -3 - 2 = -5

So the solutions are x = -1 and x = -5.