Find all complex solutions of 2x^2+3x+6=0

To calculate the roots or the solution of the equation, we can use the quadratic formula:

x = [- b ± sqrt (b^2 – 4 a c)] / 2 a

where the coefficients are:

a = 2

b = 3

c = 6

Therefore substituting this into the equation:

x = [- 3 ± sqrt (3^2 – 4 * 2 * 6)] / 2 (2)

x = -0.75 ± [sqrt (9 – 48)] / 4

x = -0.75 ± [sqrt (-39)] / 4

x = -0.75 ± 0.25 i * sqrt (39)

x = -0.75 ± 0.25 i * (6.24)

x = -0.75 ± 1.56 i

Therefore the complex solutions are:

x = -0.75 ± 0.25 i * sqrt (39)

or

x = -0.75 ± 1.56 i

.there can be infinite values of m and n that can satisfy the above equation. for each value of m of m or n. there maybe infinite sets of solution for the above

step-by-step explanation: