Is it possible to factor $x^2-6x+7$ over $\mathbb{R}$

Question

None of the online calculators seem to give me an answer.

I am trying to find the values for x. How do I do this again?

$$x^2-6x = -7$$

Then what?

Answer 1

$$ x^2-6x+7=(x^2-6x+9)-2=(x-3)^2-(\sqrt{2})^2=(x-3-\sqrt{2})(x-3+\sqrt{2}). $$

Answer 2

You can factor it over $\mathbb R$ as $(x-3-\sqrt2)(x-3+\sqrt2)$!

Answer 3

Yes, it is possible, you know this since the discriminant is greater than $0$: $$\Delta=b^2-4ac=36-4\cdot 1\cdot 7=8,$$ being the coefficients $a=1$, $b=-6$, $c=7$. Follows that $\sqrt{\Delta}=2\sqrt{2}$ and $$x_{1,2}=\frac{-b\pm\sqrt{\Delta}}{2a}=3\pm\sqrt{2}.$$ Now, $$ \begin{align} x^2-6x+7&=(x-x_1)(x-x_2)=\\ &=(x-(3+\sqrt{2}))(x-(3-\sqrt{2}))=\\ &=(x-3-\sqrt{2})(x-3+\sqrt{2}).\\ \end{align} $$