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(B) Given equation: $4x^{2} + 4x = 15$Rearranging the equation to standard form $ax^{2} + bx + c = 0$:$4x^{2} + 4x - 15 = 0$To factorize,we need two n

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

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(B) Given equation: $4x^{2} + 4x = 15$Rearranging the equation to standard form $ax^{2} + bx + c = 0$:$4x^{2} + 4x - 15 = 0$To factorize,we need two numbers whose product is $4 	imes (-15) = -60$ and whose sum is $4$. These numbers are $10$ and $-6$.Splitting the middle term:$4x^{2} + 10x - 6x - 15 = 0$Grouping the terms:$2x(2x + 5) - 3(2x + 5) = 0$Taking out the common factor:$(2x + 5)(2x - 3) = 0$Setting each factor to zero:$2x + 5 = 0$ or $2x - 3 = 0$Solving for $x$:$2x = -5 \implies x = -\frac{5}{2}$$2x = 3 \implies x = \frac{3}{2}$Thus,the solution set is $\left\{-\frac{5}{2}, \frac{3}{2}ight\}$.

Solve the following equation using the method of factorization and write its solution set: $4x^{2} + 4x = 15$.

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