Question: In $(X,d)$ metric space, If $\{x\_n\}\_{n\in\Bbb N}$ and $\{y\_n\}\_{n\in\Bbb N}$ are two sequences which converge to the same point, then prove that

$$\lim\_{n\to\infty} d(x\_n,y\_n)=0$$

This question is found in the chapter of completeness, but I could not figure out where to use completeness for this question.

Answer: 
$$6x^2-19x+10=0$$
$$6x^2-15x-4x+10=0$$
$$3x(2x-5)-2(2x-5)=0$$
$$(3x-2)(2x-5)=0$$

Hope it helps?

For your second format, you can write it as:

$$10x^2-5x-4x+2=0$$
and proceed
We can use the [Completing the Square Method](http://www.bbc.co.uk/schools/gcsebitesize/maths/algebra/quadequationshirev2.shtml) for this:

\begin{align}6x^2-19x+10&=0\\
x^2-\frac{19}{6}x+\frac{10}{6}&=0\\
\left(x-\frac{\frac{19}{6}}{2}\right)^2-\left(\frac{\frac{19}{6}}{2}\right)^2+\frac{10}{6}&=0\\
\left(x-\frac{19}{12}\right)^2-\frac{361}{144}+\frac{10}{6}&=0\\
\left(x-\frac{19}{12}\right)^2-\frac{121}{144}&=0\\
\left(x-\frac{19}{12}\right)^2&=\frac{121}{144}\\
x-\frac{19}{12}&=\pm\sqrt{\frac{121}{144}}\\
x-\frac{19}{12}&=\pm\frac{11}{12}\\
x&=\frac{19}{12}\pm\frac{11}{12}
\end{align}

And therefore we have \begin{align}x&=\frac{19}{12}+\frac{11}{12}\\
&=\frac 52\end{align}

or \begin{align}x&=\frac{19}{12}-\frac{11}{12}\\
&=\frac 23\end{align}

And so we can say that \begin{align}6x^2-19x+10&=\left(x-\frac 52\right)\left(x-\frac 23\right)\\
&=(2x-5)(3x-2)\end{align}

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**Completing the Square**

If we have an equation of the form $x^2+ax+b$, we can rewrite it as $$\left(x+\frac a2\right)^2-\left(\frac a2\right)^2+b$$