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`6x^2-13+6 = 0` |
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Answer» Here `a = 6,b = -13` and `c=6` `: . x = (-b +-sqrt b^2 -4ac)/(2a) = (+12+-sqrt(-13)^(2) - 4xx6xx6)/(2xx6)` or `x= (13+-sqrt169 -144)/(12) = (13+-5)/(12) = (18)/(12), (8)/(12)`, i.e., `x= (3)/(2), (3)/(2)` These are the roots of the given equation. |
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