Question

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

Answer 1

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`