(х+2)²+2(х+2)-3=0

To solve the equation (х+2)²+2(х+2)-3=0, we can start by simplifying it.

First, let's expand the squared term (х+2)² to get rid of the square:

(х+2)² = (х+2)(х+2) = х² + 2х + 2х + 4 = х² + 4х + 4

Now let's substitute this back into the equation:

(х+2)² + 2(х+2) - 3 = 0 (х² + 4х + 4) + 2(х+2) - 3 = 0 х² + 4х + 4 + 2х + 4 - 3 = 0 х² + 6х + 5 = 0

Now, we have a quadratic equation in standard form: ax² + bx + c = 0, where a = 1, b = 6, and c = 5. We can solve this quadratic equation by factoring, completing the square, or using the quadratic formula. Let's solve it by factoring:

(x + 1)(x + 5) = 0

To find the values of x, we set each factor equal to zero:

x + 1 = 0 or x + 5 = 0

Solving these equations separately, we get:

x = -1 or x = -5

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