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(B) The given quadratic equation is $6x^{2}-5x-1=0$.Comparing this with the standard form $ax^{2}+bx+c=0$,we get $a=6$,$b=-5$,and $c=-1$.The discrimin

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$1, -\frac{1}{6}$

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(B) The given quadratic equation is $6x^{2}-5x-1=0$.Comparing this with the standard form $ax^{2}+bx+c=0$,we get $a=6$,$b=-5$,and $c=-1$.The discriminant $D$ is given by $D = b^{2}-4ac$.$D = (-5)^{2} - 4(6)(-1) = 25 + 24 = 49$.Since $D > 0$,the roots are real and distinct.The quadratic formula is $x = \frac{-b \pm \sqrt{D}}{2a}$.$x = \frac{-(-5) \pm \sqrt{49}}{2(6)} = \frac{5 \pm 7}{12}$.Case $1$: $x = \frac{5+7}{12} = \frac{12}{12} = 1$.Case $2$: $x = \frac{5-7}{12} = \frac{-2}{12} = -\frac{1}{6}$.Thus,the solutions are $1$ and $-\frac{1}{6}$.

Solve the following equation using the quadratic formula,if the equation has a solution in $R$: $6x^{2}-5x-1=0$.

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