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

i think the correct option is the second option.

step-by-step explanation:

we can use this expression to calculate the energy gain as a particle moves with a field or the energy loss as it moves against it. (note that it is also used to define the electron-volt, a unit of energy equal to that gained by an electron as it falls through a pd of 1 v, i.e. w = 1 ev = 1.6 10-19 c 1 v = 1.6 10-19j).in fact this expression is really nothing more than the statement that potential is the potential energy per unit charge. in this case, v= w/qis simply stating that the change in potential energy(or work done) per unit charge is equal to the change in potential (or potential difference), v.student questions: uniform electric fields2electron beamcathode+5 kvanode0 v