How do you find all the real and complex roots of #3x^2 - x + 2 = 0#?

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Answer 1 · 2018-08-08T09:53:47

#x=(1+sqrt23*i)/6 or x=(1-sqrt23*i)/6#

Here ,

#3x^2-x+2=0#

Comparing with #ax^2+bx+c=0#

#a=3,b=-1 and c=2#

#:.Delta=b^2-4ac=(-1)^2-4(3)(2)#

#:.color(red)(Delta=1-24=-23 < 0=>ul(x !inRR and x inCC)#

#:.Delta=23(-1)#

#:.Delta=23i^2to[because i^2=-1]#

#:.sqrtDelta=sqrt23*i#

So,

#x=(-b+-sqrtDelta)/(2a)=(1+-sqrt23*i)/(2xx3)#

#:.x=(1+-sqrt23*i)/6#

Hence ,

#x=(1+sqrt23*i)/6 or x=(1-sqrt23*i)/6#