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(B) The given equation is $-3x^{2} + 5x + 12 = 0$.Comparing this with the standard form $ax^{2} + bx + c = 0$,we get $a = -3$,$b = 5$,and $c = 12$.The

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$-\frac{4}{3}, 3$

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(B) The given equation is $-3x^{2} + 5x + 12 = 0$.Comparing this with the standard form $ax^{2} + bx + c = 0$,we get $a = -3$,$b = 5$,and $c = 12$.The quadratic formula is given by $x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}$.Substituting the values,we get $x = \frac{-(5) \pm \sqrt{(5)^{2} - 4(-3)(12)}}{2(-3)}$.Simplifying the expression,$x = \frac{-5 \pm \sqrt{25 + 144}}{-6} = \frac{-5 \pm \sqrt{169}}{-6}$.Since $\sqrt{169} = 13$,we have $x = \frac{-5 \pm 13}{-6}$.Taking the positive sign: $x = \frac{-5 + 13}{-6} = \frac{8}{-6} = -\frac{4}{3}$.Taking the negative sign: $x = \frac{-5 - 13}{-6} = \frac{-18}{-6} = 3$.Thus,the roots of the equation are $-\frac{4}{3}$ and $3$.

Find the roots of the quadratic equation by using the quadratic formula:
$-3x^{2} + 5x + 12 = 0$

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Find the roots of the quadratic equation by using the quadratic formula:$-3x^{2} + 5x + 12 = 0$