$5x^2-6x-2 = 0$ is a quadratic equation and the roots of this quadratic equation can be evaluated by the method of completing the square.

$5x^2$ is a term in the quadratic equation and the literal coefficient of $x^2$ is $5$. It should be eliminated from this term and it can be done by dividing both sides of the quadratic equation by $5$.

$\implies$ $\dfrac{5x^2-6x-2}{5}$ $\,=\,$ $\dfrac{0}{5}$

$\implies$ $\dfrac{5x^2}{5}$ $-$ $\dfrac{6x}{5}$ $-$ $\dfrac{2}{5}$ $\,=\,$ $0$

$\implies$ $\require{cancel} \dfrac{\cancel{5}x^2}{\cancel{5}}$ $-$ $\dfrac{6x}{5}$ $-$ $\dfrac{2}{5}$ $\,=\,$ $0$

$\implies$ $x^2$ $-$ $\dfrac{6x}{5}$ $-$ $\dfrac{2}{5}$ $\,=\,$ $0$

$x^2$ $-$ $\dfrac{6x}{5}$ $-$ $\dfrac{2}{5}$ $\,=\,$ $0$

The algebraic expression in the left-hand side of the equation is $x^2$ $-$ $\dfrac{6x}{5}$ $-$ $\dfrac{2}{5}$. It should be converted into square form of a binomial completely to find the roots of the quadratic equation $5x^2-6x-2 = 0$ mathematically.

$\implies$ $x^2$ $-$ $\dfrac{2 \times 3 \times x}{5}$ $-$ $\dfrac{2}{5}$ $\,=\,$ $0$

$\implies$ $x^2$ $-$ $\dfrac{2 \times x \times 3}{5}$ $-$ $\dfrac{2}{5}$ $\,=\,$ $0$

$\implies$ $x^2$ $-$ $2 \times x \times \dfrac{3}{5}$ $-$ $\dfrac{2}{5}$ $\,=\,$ $0$

$\implies$ $x^2$ $-$ $2(x)\Big(\dfrac{3}{5}\Big)$ $-$ $\dfrac{2}{5}$ $\,=\,$ $0$

The first two terms in the quadratic equation represent the expansion of the square of difference of the two terms formula but its third term is missed. So, add and subtract square of $\dfrac{3}{5}$ to the left-hand side of the quadratic equation.

$\implies$ $x^2$ $-$ $2(x)\Big(\dfrac{3}{5}\Big)$ $+$ ${\Big(\dfrac{3}{5}\Big)}^2$ $-$ ${\Big(\dfrac{3}{5}\Big)}^2$ $-$ $\dfrac{2}{5}$ $\,=\,$ $0$

$\implies$ $x^2$ $-$ $2(x)\Big(\dfrac{3}{5}\Big)$ $+$ ${\Big(\dfrac{3}{5}\Big)}^2$ $-$ $\dfrac{9}{25}$ $-$ $\dfrac{2}{5}$ $\,=\,$ $0$

$\implies$ $x^2$ $-$ $2(x)\Big(\dfrac{3}{5}\Big)$ $+$ ${\Big(\dfrac{3}{5}\Big)}^2$ $\,=\,$ $\dfrac{9}{25}$ $+$ $\dfrac{2}{5}$

The left hand side of the quadratic equation is the expansion of square of difference of two terms identity. So, the algebraic expression can be written in square form of a binomial completely.

$\implies$ ${\Big(x-\dfrac{3}{5}\Big)}^2$ $\,=\,$ $\dfrac{9}{25}$ $+$ $\dfrac{2}{5}$

Now, add the sum of the fractions in the right-hand side of the equation to get their sum.

$\implies$ ${\Big(x-\dfrac{3}{5}\Big)}^2$ $\,=\,$ $\dfrac{1 \times 9 + 5 \times 2}{25}$

$\implies$ ${\Big(x-\dfrac{3}{5}\Big)}^2$ $\,=\,$ $\dfrac{9+10}{25}$

$\implies$ ${\Big(x-\dfrac{3}{5}\Big)}^2$ $\,=\,$ $\dfrac{19}{25}$

${\Big(x-\dfrac{3}{5}\Big)}^2$ $\,=\,$ $\dfrac{19}{25}$

Take square root both sides of the equation to find the solution of the quadratic equation.

$\implies$ $x-\dfrac{3}{5}$ $\,=\,$ $\pm \sqrt{\dfrac{19}{25}}$

The equation is a linear equation in one variable. It can be solved by the transposing method.

$\implies$ $x-\dfrac{3}{5}$ $\,=\,$ $\pm \dfrac{\sqrt{19}}{5}$

$\implies$ $x$ $\,=\,$ $\dfrac{3}{5} \pm \dfrac{\sqrt{19}}{5}$

$\implies$ $x \,=\, \dfrac{3 \pm \sqrt{19}}{5}$

Therefore $x \,=\, \dfrac{3+\sqrt{19}}{5}$ and $x \,=\, \dfrac{3-\sqrt{19}}{5}$ are the roots of the quadratic equation. Thus, the quadratic equation $5x^2-6x-2 = 0$ is solved by the method of Completing the square.

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