(i+3)+i*(2i-4)=0

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Solution for (i+3)+i*(2i-4)=0 equation:


Simplifying
(i + 3) + i(2i + -4) = 0

Reorder the terms:
(3 + i) + i(2i + -4) = 0

Remove parenthesis around (3 + i)
3 + i + i(2i + -4) = 0

Reorder the terms:
3 + i + i(-4 + 2i) = 0
3 + i + (-4 * i + 2i * i) = 0
3 + i + (-4i + 2i2) = 0

Combine like terms: i + -4i = -3i
3 + -3i + 2i2 = 0

Solving
3 + -3i + 2i2 = 0

Solving for variable 'i'.

Begin completing the square.  Divide all terms by
2 the coefficient of the squared term: 

Divide each side by '2'.
1.5 + -1.5i + i2 = 0

Move the constant term to the right:

Add '-1.5' to each side of the equation.
1.5 + -1.5i + -1.5 + i2 = 0 + -1.5

Reorder the terms:
1.5 + -1.5 + -1.5i + i2 = 0 + -1.5

Combine like terms: 1.5 + -1.5 = 0.0
0.0 + -1.5i + i2 = 0 + -1.5
-1.5i + i2 = 0 + -1.5

Combine like terms: 0 + -1.5 = -1.5
-1.5i + i2 = -1.5

The i term is -1.5i.  Take half its coefficient (-0.75).
Square it (0.5625) and add it to both sides.

Add '0.5625' to each side of the equation.
-1.5i + 0.5625 + i2 = -1.5 + 0.5625

Reorder the terms:
0.5625 + -1.5i + i2 = -1.5 + 0.5625

Combine like terms: -1.5 + 0.5625 = -0.9375
0.5625 + -1.5i + i2 = -0.9375

Factor a perfect square on the left side:
(i + -0.75)(i + -0.75) = -0.9375

Can't calculate square root of the right side.

The solution to this equation could not be determined.

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