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(D) Given equation: $3x^{2} = -11x - 10$.Rearranging the terms to standard form $ax^{2} + bx + c = 0$:$3x^{2} + 11x + 10 = 0$.To factorize,we need two

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$-\frac{5}{3}, -2$

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(D) Given equation: $3x^{2} = -11x - 10$.Rearranging the terms to standard form $ax^{2} + bx + c = 0$:$3x^{2} + 11x + 10 = 0$.To factorize,we need two numbers whose product is $3 	imes 10 = 30$ and whose sum is $11$.These numbers are $6$ and $5$.Splitting the middle term:$3x^{2} + 6x + 5x + 10 = 0$.Grouping the terms:$3x(x + 2) + 5(x + 2) = 0$.$(3x + 5)(x + 2) = 0$.Setting each factor to zero:$3x + 5 = 0 \implies x = -\frac{5}{3}$.$x + 2 = 0 \implies x = -2$.Therefore,the solutions are $x = -\frac{5}{3}, -2$.

Solve the following quadratic equation using the method of factorization: $3x^{2} = -11x - 10$.

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