Factorise it.urgently required.
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here is the right answer
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by quadratic formula
b^2-4ac
=(-10)^2-4×3×(-8)
=100+32=132
since b^2-4ac>0
it has two distinct roots
x=(-b plusminus √b^2-4ac)/2a
=[-(-10)plusminus√132]/2×3
=[10plusminus2√33]/6
=2[5plusminus√33]/6
=[5plusminus√33]/3
so X=(5+√33)/3and (5-√33)/3
Hope it will be helpful.
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