Find the roots of the equation $5x^{2}-6x-2=0$ by the method of completing the square.
(N/A) Multiplying the equation throughout by $5$,we get:
$25x^{2}-30x-10=0$
This can be rewritten as:
$(5x)^{2}-2 \times (5x) \times 3 + 3^{2} - 3^{2} - 10 = 0$
$(5x-3)^{2} - 9 - 10 = 0$
$(5x-3)^{2} - 19 = 0$
$(5x-3)^{2} = 19$
Taking the square root on both sides:
$5x-3 = \pm \sqrt{19}$
$5x = 3 \pm \sqrt{19}$
$x = \frac{3 \pm \sqrt{19}}{5}$
Therefore,the roots are $\frac{3+\sqrt{19}}{5}$ and $\frac{3-\sqrt{19}}{5}$.
$25x^{2}-30x-10=0$
This can be rewritten as:
$(5x)^{2}-2 \times (5x) \times 3 + 3^{2} - 3^{2} - 10 = 0$
$(5x-3)^{2} - 9 - 10 = 0$
$(5x-3)^{2} - 19 = 0$
$(5x-3)^{2} = 19$
Taking the square root on both sides:
$5x-3 = \pm \sqrt{19}$
$5x = 3 \pm \sqrt{19}$
$x = \frac{3 \pm \sqrt{19}}{5}$
Therefore,the roots are $\frac{3+\sqrt{19}}{5}$ and $\frac{3-\sqrt{19}}{5}$.
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