(х+2)²+2(х+2)-3=0

To solve the equation (х+2)²+2(х+2)-3=0, we can first simplify it:

(x+2)²+2(x+2)-3 = 0

Expanding the square term (x+2)² results in:

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

(x²+4x+4)+2(x+2)-3 = 0

x²+4x+4+2x+4-3 = 0

x²+6x+5 = 0

Now, we can use the quadratic formula to solve for x. The quadratic formula states that for a quadratic equation of the form ax²+bx+c=0, the solutions for x can be found using:

x = (-b ± √(b²-4ac)) / (2a)

In our equation, a=1, b=6, and c=5. Plugging these values into the quadratic formula, we have:

x = (-6 ± √(6²-4(1)(5))) / (2(1))

Simplifying,

x = (-6 ± √(36-20)) / 2

x = (-6 ± √(16)) / 2

x = (-6 ± 4) / 2

x1 = (-6 + 4) / 2 = -1

x2 = (-6 - 4) / 2 = -5

Therefore, the solutions to the equation (х+2)²+2(х+2)-3=0 are x = -1 and x = -5.