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Yes.
But note that if b2 - 4ac is negative, there are no real solutions to the quadratic equation to be found.
When complex numbers are used, this is not a problem as sqrt(-1) = i and so if b2 - 4ac is negative, "sqrt(b2 - 4ac)" becomes "i sqrt(4ac - b2)", meaning the solutions are:
x = -b/2a + i/2a sqrt(4ac-b2)
x = -b/2a - i/2a sqrt(4ac-b2)
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