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 ,

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

How do you find all the real and complex roots of 3x^2 - x + 2 = 03x^2 - x + 2 = 0?