$2x^2+x-4 = 0$ is a quadratic equation and solution of this equation has to be calculated by the completing the square method in this problem.

The number $-4$ is a constant in the given quadratic equation and shift it to right-hand side of the quadratic equation.

$2x^2+x-4 = 0$

$\implies$ $2x^2+x = 4$

The literal coefficient of $x^2$ in the first term is $2$ and divide both sides of this mathematical equation by this number.

$\implies$ $\dfrac{2x^2+x}{2}$ $\,=\,$ $\dfrac{4}{2}$

$\implies$ $\dfrac{2x^2}{2}+\dfrac{x}{2}$ $\,=\,$ $\dfrac{4}{2}$

$\implies$ $\require{cancel} \dfrac{\cancel{2}x^2}{\cancel{2}}+\dfrac{x}{2}$ $\,=\,$ $\require{cancel} \dfrac{\cancel{4}}{\cancel{2}}$

$\implies$ $x^2+\dfrac{x}{2}$ $\,=\,$ $2$

Observe both sides of the mathematical equation. One side of the equation is in algebraic form and other side is a constant. The signs of both terms in the algebraic expression are positive. So, the algebraic expression can be converted in square form as per square of sum of two terms formula.

$\implies$ $x^2+1 \times \dfrac{x}{2}$ $\,=\,$ $2$

$\implies$ $x^2+\dfrac{2}{2} \times \dfrac{x}{2}$ $\,=\,$ $2$

$\implies$ $x^2+2 \times \dfrac{1}{2} \times x \times \dfrac{1}{2}$ $\,=\,$ $2$

$\implies$ $x^2+2 \times x \times \dfrac{1}{2} \times \dfrac{1}{2}$ $\,=\,$ $2$

$\implies$ $x^2+2 \times x \times \dfrac{1 \times 1}{2 \times 2}$ $\,=\,$ $2$

$\implies$ $x^2+2 \times x \times \dfrac{1}{4}$ $\,=\,$ $2$

$\implies$ $x^2+2x\Big(\dfrac{1}{4}\Big)$ $\,=\,$ $2$

The algebraic expression is almost appeared same as the expansion of square of sum of two terms formula but a term is missed. So, add and subtract that term for expressing it in the square form completely.

$\implies$ $x^2+2x\Big(\dfrac{1}{4}\Big)+{\Big(\dfrac{1}{4}\Big)}^2-{\Big(\dfrac{1}{4}\Big)}^2$ $\,=\,$ $2$

$\implies$ $x^2+2x\Big(\dfrac{1}{4}\Big)+{\Big(\dfrac{1}{4}\Big)}^2$ $\,=\,$ $2+{\Big(\dfrac{1}{4}\Big)}^2$

$\implies$ $x^2+2x\Big(\dfrac{1}{4}\Big)+{\Big(\dfrac{1}{4}\Big)}^2$ $\,=\,$ $2+\dfrac{1^2}{4^2}$

$\implies$ $x^2+2x\Big(\dfrac{1}{4}\Big)+{\Big(\dfrac{1}{4}\Big)}^2$ $\,=\,$ $2+\dfrac{1}{16}$

$\implies$ $x^2+2x\Big(\dfrac{1}{4}\Big)+{\Big(\dfrac{1}{4}\Big)}^2$ $\,=\,$ $\dfrac{2 \times 16 + 1}{16}$

$\implies$ $x^2+2x\Big(\dfrac{1}{4}\Big)+{\Big(\dfrac{1}{4}\Big)}^2$ $\,=\,$ $\dfrac{32+1}{16}$

$\implies$ $x^2+2x\Big(\dfrac{1}{4}\Big)+{\Big(\dfrac{1}{4}\Big)}^2$ $\,=\,$ $\dfrac{33}{16}$

$\implies$ ${\Big(x+\dfrac{1}{4}\Big)}^2$ $\,=\,$ $\dfrac{33}{16}$

$\implies$ $x+\dfrac{1}{4}$ $\,=\,$ $\pm \sqrt{\dfrac{33}{16}}$

$\implies$ $x+\dfrac{1}{4}$ $\,=\,$ $\pm \dfrac{\sqrt{33}}{4}$

$\implies$ $x$ $\,=\,$ $-\dfrac{1}{4} \pm \dfrac{\sqrt{33}}{4}$

$\implies$ $x$ $\,=\,$ $\dfrac{-1 \pm \sqrt{33}}{4}$

$\,\,\, \therefore \,\,\,\,\,\,$ $x \,=\, \dfrac{-1+\sqrt{33}}{4}$ and $x \,=\, \dfrac{-1-\sqrt{33}}{4}$

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