How do you solve x^2 + 3x = -4 ?

1 Answer
Nghi N.
Apr 29, 2016

There are no real roots.

Explanation:

y = x^2 + 3x + 4 = 0.
D = d^2 = b^2 - 4ac = 9 - 16 = - 7
Since D < 0, there are no real roots. There are 2 complex roots.
d^2 = -7 = 7i^2 --> d = +- isqrt7
x = -b/(2a) +- d/(2a) = -3/2 +- isqrt7/2 = (-3 +- isqrt7)/2

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