silemus.

38Quadratic with complex roots

Solve x2+4x+13=0\displaystyle x^2 + 4x + 13 = 0.

The equation has no real solutions, only complex ones: x = -2 ± 3i. Using the quadratic formula with a=1, b=4, c=13 gives √(16-52) = √-36 = 6i, so x = (-4 ± 6i)/2 = -2 ± 3i. Checking x = -2+3i back into the equation gives 0.

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