$2x^2-9x+10$ $\,=\,$ $0$ Is a given quadratic equation in this problem and it should be solved by the factorization method.

For using the factorisation method, the quadratic equation should be in descending order. It is already given in descending order. Similarly, all the terms in the left-hand side of the equation and there is no term in the right hand side of the equation. So, we can start the process for factoring the quadratic equation to solve it.

$2x^2$ is a term in which the variable $x$ consists of $2$ as exponent and $10$ is a constant term in the left hand side of the quadratic equation $2x^2-9x+10 \,=\, 0$.

Now, multiply the first term $2x^2$ by the constant term $10$ for calculating their product.

$\implies$ $2x^2 \times 10 \,=\, 20x^2$

$-9x$ is a middle term in the given quadratic equation $2x^2-9x+10 \,=\, 0$. The term $-9x$ should be split as either sum or difference of two terms but their product should be equal to the $20x^2$.

The signs of the term $2x^2$ and the constant term $10$ are positive. Hence, the sign of their product $20x^2$ is also positive but the sign of the middle term $-9x$ is negative. So, the term $-9x$ should be split as the sum of two negative terms. The following is the list of possible ways to split the term $-9x$ as the sum of two terms.

$(1).\,\,\,$ $-x-8x \,=\, -9x$

$(2).\,\,\,$ $-2x-7x \,=\, -9x$

$(3).\,\,\,$ $-3x-6x \,=\, -9x$

$(4).\,\,\,$ $-4x-5x \,=\, -9x$

The product of $-x$ and $-8x$ is not equal to $20x^2$. The product of $-2x$ and $-7x$ is also not equal to $20x^2$. Similarly, the product of $-3x$ and $-6x$ is also not equal to $20x^2$ but the product of $-4x$ and $-5x$ is equal to the $20x^2$. Hence, the term $-9x$ should be split as the sum of the terms $-4x$ and $-5x$ in the quadratic equation $2x^2-9x+10 \,=\, 0$.

$\implies$ $2x^2-4x-5x+10$ $\,=\,$ $0$

The trinomial in the quadratic expression is expanded as a quadrinomial. There is a factor in every two terms commonly in the quadrinomial.

$\implies$ $2x \times x$ $-$ $2x \times 2$ $-$ $5 \times x$ $+$ $5 \times 2$ $\,=\,$ $0$

Now, take the common factor out from each two terms in the quadratic equation.

$\implies$ $2x \times (x-2)$ $-$ $5 \times x$ $+$ $5 \times 2$ $\,=\,$ $0$

$\implies$ $2x \times (x-2)$ $-$ $5 \times x$ $+$ $(-5) \times (-2)$ $\,=\,$ $0$

$\implies$ $2x \times (x-2)$ $-$ $5 \times x$ $-$ $5 \times (-2)$ $\,=\,$ $0$

$\implies$ $2x \times (x-2)$ $-$ $5 \times (x-2)$ $\,=\,$ $0$

$\implies$ $(x-2)(2x-5)$ $\,=\,$ $0$

The given quadratic equation $2x^2-9x+10 \,=\, 0$ is successfully factored as $(x-2)(2x-5) = 0$. Now, the zero product property can be used to solve the quadratic equation by making each linear expression in one variable equals to zero.

$\implies$ $x-2 \,=\, 0$ or $2x-5 \,=\, 0$

$\implies$ $x \,=\, 2$ or $2x \,=\, 5$

$\,\,\,\therefore\,\,\,\,\,\,$ $x \,=\, 2$ or $x \,=\, \dfrac{5}{2}$

Therefore, the solution set for the quadratic equation $2x^2-9x+10 \,=\, 0$ by the factorization method is $\bigg\{2, \dfrac{5}{2}\bigg\}$

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