The two times $x$ squared minus $x$ minus six is equal to zero, is a given quadratic equation in this problem.

$2x^2-x-6$ $\,=\,$ $0$

It is mentioned that the $2$ times $x$ square minus $x$ minus $6$ is equal to $0$ should be solved by the factorization or factorisation method. So, let us learn how to solve $2$ times square of $x$ minus $x$ minus $6$ is equal to zero by factorisation.

The two times $x$ squared is a first term and $-6$ is a constant term on the left hand side of the given equation. Now, multiply both of them to find their product.

$\implies$ $2x^2$ $\times$ $(-6)$ $\,=\,$ $-12x^2$

Now, split the product $-12x^2$ as two factors in such a way that the sum or difference of the factors should be exactly equal to the middle term $-x$.

$(1).\,\,$ $-12x^2$ $\,=\,$ $(-12x) \times x$ but $-12x+x$ $\,\ne\,$ $-x$ and $-12x-x$ $\,\ne\,$ $-x$

It is proved that the sum or difference of the factors $-12x$ and $x$ are not equal to the middle term $-x$.

$(2).\,\,$ $-12x^2$ $\,=\,$ $(-6x) \times 2x$ but $-6x+2x$ $\,\ne\,$ $-x$ and $-6x-2x$ $\,\ne\,$ $-x$

It is also proved that the sum or difference of the factors $-6x$ and $2x$ are not equal to the middle term $-x$.

$(3).\,\,$ $-12x^2$ $\,=\,$ $(-4x) \times 3x$ but $-4x+3x$ $\,=\,$ $-x$ and $-4x-3x$ $\,\ne\,$ $-x$

Therefore, it is proved that the sum of the factors $-4x$ and $3x$ is equal to the middle term $-x$. Hence, the middle $-x$ in the quadratic expression of the equation can be split as the sum of $-4x$ and $3x$.

$\implies$ $2x^2-4x+3x-6$ $\,=\,$ $0$

The quadratic expression $2$ times $x$ squared minus $x$ minus $6$ on the left hand side of the equation is expanded as a polynomial in four terms by splitting the middle term. Now, the quadratic expression in the equation can be arranged as two groups as follows for factoring.

$\implies$ $(2x^2-4x)+(3x-6)$ $\,=\,$ $0$

Now, split the terms in each group in such a way that the terms in each group have a common factor.

$\implies$ $(2x \times x-2 \times 2x)$ $+$ $(3 \times x-3 \times 2)$ $\,=\,$ $0$

The two times $x$ is a common factor in the terms of first group and $3$ is a common factor in the two terms of the second group. So, take the common factor out from the terms in every group of the left hand side of the equation.

$\implies$ $2x \times (x-2)$ $+$ $3 \times (x-2)$ $\,=\,$ $0$

Finally, take the factor $x-2$ common from the terms for finishing the process of factoring by grouping method.

$\implies$ $(x-2) \times (2x+3)$ $\,=\,$ $0$

$\implies$ $(x-2)(2x+3)$ $\,=\,$ $0$

The quadratic expression $2$ times $x$ squared minus $x$ minus $6$ is successfully factored as a product of two binomials $x$ minus $2$ and $2$ times $x$ plus $3$. Now, make linear expressions in one variable $x-2$ and $2x+3$ are equal to zero to find the solutions.

$\implies$ $x-2 \,=\, 0$ or $2x+3 \,=\, 0$

Now, simplify each linear equation in one variable and find the value of $x$.

$\implies$ $x \,=\, 2$ or $2x \,=\, -3$

$\implies$ $x \,=\, 2$ or $x \,=\, \dfrac{-3}{2}$

$\,\,\,\therefore\,\,\,\,\,\,$ $x \,=\, 2$ or $x \,=\, -\dfrac{3}{2}$

The solution set of the given quadratic equation $2x^2-x-6 = 0$ is $\bigg\{-\dfrac{3}{2}, 2\bigg\}$

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