z^2-(3+i)z+4=0

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


Simplifying
z2 + -1(3 + i) * z + 4 = 0

Reorder the terms for easier multiplication:
z2 + -1z(3 + i) + 4 = 0
z2 + (3 * -1z + i * -1z) + 4 = 0

Reorder the terms:
z2 + (-1iz + -3z) + 4 = 0
z2 + (-1iz + -3z) + 4 = 0

Reorder the terms:
4 + -1iz + -3z + z2 = 0

Solving
4 + -1iz + -3z + z2 = 0

Solving for variable 'i'.

Move all terms containing i to the left, all other terms to the right.

Add '-4' to each side of the equation.
4 + -1iz + -3z + -4 + z2 = 0 + -4

Reorder the terms:
4 + -4 + -1iz + -3z + z2 = 0 + -4

Combine like terms: 4 + -4 = 0
0 + -1iz + -3z + z2 = 0 + -4
-1iz + -3z + z2 = 0 + -4

Combine like terms: 0 + -4 = -4
-1iz + -3z + z2 = -4

Add '3z' to each side of the equation.
-1iz + -3z + 3z + z2 = -4 + 3z

Combine like terms: -3z + 3z = 0
-1iz + 0 + z2 = -4 + 3z
-1iz + z2 = -4 + 3z

Add '-1z2' to each side of the equation.
-1iz + z2 + -1z2 = -4 + 3z + -1z2

Combine like terms: z2 + -1z2 = 0
-1iz + 0 = -4 + 3z + -1z2
-1iz = -4 + 3z + -1z2

Divide each side by '-1z'.
i = 4z-1 + -3 + z

Simplifying
i = 4z-1 + -3 + z

Reorder the terms:
i = -3 + 4z-1 + z

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