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

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Solution for (x^2)+(4x)+1=0 equation:


Simplifying
(x2) + (4x) + 1 = 0
x2 + (4x) + 1 = 0

Reorder the terms:
1 + (4x) + x2 = 0

Solving
1 + (4x) + x2 = 0

Solving for variable 'x'.

Begin completing the square.

Move the constant term to the right:

Add '-1' to each side of the equation.
1 + (4x) + -1 + x2 = 0 + -1

Reorder the terms:
1 + -1 + (4x) + x2 = 0 + -1

Combine like terms: 1 + -1 = 0
0 + (4x) + x2 = 0 + -1
(4x) + x2 = 0 + -1

Combine like terms: 0 + -1 = -1
(4x) + x2 = -1

The x term is (4x).  Take half its coefficient (2).
Square it (4) and add it to both sides.

Add '4' to each side of the equation.
(4x) + 4 + x2 = -1 + 4

Reorder the terms:
4 + (4x) + x2 = -1 + 4

Combine like terms: -1 + 4 = 3
4 + (4x) + x2 = 3

Factor a perfect square on the left side:
(x + 2)(x + 2) = 3

Calculate the square root of the right side: 1.732050808

Break this problem into two subproblems by setting 
(x + 2) equal to 1.732050808 and -1.732050808.

Subproblem 1

x + 2 = 1.732050808 Simplifying x + 2 = 1.732050808 Reorder the terms: 2 + x = 1.732050808 Solving 2 + x = 1.732050808 Solving for variable 'x'. Move all terms containing x to the left, all other terms to the right. Add '-2' to each side of the equation. 2 + -2 + x = 1.732050808 + -2 Combine like terms: 2 + -2 = 0 0 + x = 1.732050808 + -2 x = 1.732050808 + -2 Combine like terms: 1.732050808 + -2 = -0.267949192 x = -0.267949192 Simplifying x = -0.267949192

Subproblem 2

x + 2 = -1.732050808 Simplifying x + 2 = -1.732050808 Reorder the terms: 2 + x = -1.732050808 Solving 2 + x = -1.732050808 Solving for variable 'x'. Move all terms containing x to the left, all other terms to the right. Add '-2' to each side of the equation. 2 + -2 + x = -1.732050808 + -2 Combine like terms: 2 + -2 = 0 0 + x = -1.732050808 + -2 x = -1.732050808 + -2 Combine like terms: -1.732050808 + -2 = -3.732050808 x = -3.732050808 Simplifying x = -3.732050808

Solution

The solution to the problem is based on the solutions from the subproblems. x = {-0.267949192, -3.732050808}

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