(x^2)+(2x)-160=0

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Solution for (x^2)+(2x)-160=0 equation:


Simplifying
(x2) + (2x) + -160 = 0
x2 + (2x) + -160 = 0

Reorder the terms:
-160 + (2x) + x2 = 0

Solving
-160 + (2x) + x2 = 0

Solving for variable 'x'.

Begin completing the square.

Move the constant term to the right:

Add '160' to each side of the equation.
-160 + (2x) + 160 + x2 = 0 + 160

Reorder the terms:
-160 + 160 + (2x) + x2 = 0 + 160

Combine like terms: -160 + 160 = 0
0 + (2x) + x2 = 0 + 160
(2x) + x2 = 0 + 160

Combine like terms: 0 + 160 = 160
(2x) + x2 = 160

The x term is (2x).  Take half its coefficient (1).
Square it (1) and add it to both sides.

Add '1' to each side of the equation.
(2x) + 1 + x2 = 160 + 1

Reorder the terms:
1 + (2x) + x2 = 160 + 1

Combine like terms: 160 + 1 = 161
1 + (2x) + x2 = 161

Factor a perfect square on the left side:
(x + 1)(x + 1) = 161

Calculate the square root of the right side: 12.68857754

Break this problem into two subproblems by setting 
(x + 1) equal to 12.68857754 and -12.68857754.

Subproblem 1

x + 1 = 12.68857754 Simplifying x + 1 = 12.68857754 Reorder the terms: 1 + x = 12.68857754 Solving 1 + x = 12.68857754 Solving for variable 'x'. Move all terms containing x to the left, all other terms to the right. Add '-1' to each side of the equation. 1 + -1 + x = 12.68857754 + -1 Combine like terms: 1 + -1 = 0 0 + x = 12.68857754 + -1 x = 12.68857754 + -1 Combine like terms: 12.68857754 + -1 = 11.68857754 x = 11.68857754 Simplifying x = 11.68857754

Subproblem 2

x + 1 = -12.68857754 Simplifying x + 1 = -12.68857754 Reorder the terms: 1 + x = -12.68857754 Solving 1 + x = -12.68857754 Solving for variable 'x'. Move all terms containing x to the left, all other terms to the right. Add '-1' to each side of the equation. 1 + -1 + x = -12.68857754 + -1 Combine like terms: 1 + -1 = 0 0 + x = -12.68857754 + -1 x = -12.68857754 + -1 Combine like terms: -12.68857754 + -1 = -13.68857754 x = -13.68857754 Simplifying x = -13.68857754

Solution

The solution to the problem is based on the solutions from the subproblems. x = {11.68857754, -13.68857754}

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