v^2+14*v+14=0

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Solution for v^2+14*v+14=0 equation:


Simplifying
v2 + 14v + 14 = 0

Reorder the terms:
14 + 14v + v2 = 0

Solving
14 + 14v + v2 = 0

Solving for variable 'v'.

Begin completing the square.

Move the constant term to the right:

Add '-14' to each side of the equation.
14 + 14v + -14 + v2 = 0 + -14

Reorder the terms:
14 + -14 + 14v + v2 = 0 + -14

Combine like terms: 14 + -14 = 0
0 + 14v + v2 = 0 + -14
14v + v2 = 0 + -14

Combine like terms: 0 + -14 = -14
14v + v2 = -14

The v term is 14v.  Take half its coefficient (7).
Square it (49) and add it to both sides.

Add '49' to each side of the equation.
14v + 49 + v2 = -14 + 49

Reorder the terms:
49 + 14v + v2 = -14 + 49

Combine like terms: -14 + 49 = 35
49 + 14v + v2 = 35

Factor a perfect square on the left side:
(v + 7)(v + 7) = 35

Calculate the square root of the right side: 5.916079783

Break this problem into two subproblems by setting 
(v + 7) equal to 5.916079783 and -5.916079783.

Subproblem 1

v + 7 = 5.916079783 Simplifying v + 7 = 5.916079783 Reorder the terms: 7 + v = 5.916079783 Solving 7 + v = 5.916079783 Solving for variable 'v'. Move all terms containing v to the left, all other terms to the right. Add '-7' to each side of the equation. 7 + -7 + v = 5.916079783 + -7 Combine like terms: 7 + -7 = 0 0 + v = 5.916079783 + -7 v = 5.916079783 + -7 Combine like terms: 5.916079783 + -7 = -1.083920217 v = -1.083920217 Simplifying v = -1.083920217

Subproblem 2

v + 7 = -5.916079783 Simplifying v + 7 = -5.916079783 Reorder the terms: 7 + v = -5.916079783 Solving 7 + v = -5.916079783 Solving for variable 'v'. Move all terms containing v to the left, all other terms to the right. Add '-7' to each side of the equation. 7 + -7 + v = -5.916079783 + -7 Combine like terms: 7 + -7 = 0 0 + v = -5.916079783 + -7 v = -5.916079783 + -7 Combine like terms: -5.916079783 + -7 = -12.916079783 v = -12.916079783 Simplifying v = -12.916079783

Solution

The solution to the problem is based on the solutions from the subproblems. v = {-1.083920217, -12.916079783}

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