How do you find the solution set in a+bi form of: 3x^2 - x + 1 = 0 ?

Answer 1

.use the quadratic formula

x = (-b+/- sqrt (b^2-4ac) ) / (2a)

where your a=3 b=-1 c=1

x = (+1 +/- sqrt(1-4(3)(1)) / (2*3)
=(1 +/- sqrt(-11) ) /6

sqrt (-11) = i * sqrt(11)

x = 1/6 +/- sqrt(11) / 6 * i

:)

Answer 2

3x^2 - x + 1 = 0

x = (-b ± sqrt(b^2 - 4ac))/(2a)

x = (-(-1) ± sqrt((-1)^2 - 4(3)(1)))/(2(3))
x = (1 ± sqrt(1 - 12))/6
x = (1 ± sqrt(-11))/6
x = (1 ± isqrt(11))/6
x = (1/6)(1 ± isqrt(11))

Answer 3

x=((1+-sqrt(1-12)/6
x=1/6+1/6sqrt(11)*i
x=1/6-1/6sqrt(11)*i