# Equation .Solve the equation by completing the square x^2 +2x+5=0 .

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We have to solve x^2 +2x+5=0 by completion of squares.

We use the relation x^2 + 2xy + y^2 = (x + y)^2

x^2 +2x+5=0

=> x^2 + 2x*1 + 1^2 + 4 = 0

=> (x + 1)^2 + 4 = 0

=> (x + 1)^2 = -4

=> x + 1 = 2i and x + 1 = -2i

=> x = -1 + 2i and x = -1 - 2i

**The solution of the equation is x = -1 + 2i and x = -1 - 2i**

We'll begin by subtracting 5 both sides, to shift the number alone on the right side of the equation, so that becoming more clear what we have to add to the left side to complete the square.

x^2 +2x = -5

We'll complete the square by adding the number 1 to both side, to get a perfect square to the left side.

x^2 +2x + 1 = -5 + 1

We'll write the left side as a perfect square:

(x + 1)^2 = -4

x + 1 = sqrt -4

x + 1 = 2i

x1 = -1 + 2i

x2 = -1 - 2i

**The complex solutions of the equation are: { -1 + 2i ; -1 - 2i}.**