Solving quadratic equation $x^2-(k+1)x+k+1=0$ using quadratic formula Disclaimer
I am new here so please mind how i have  executed this question. Feel free to comment on how to improve as i have read on “how to ask a good question”.
Problem
I am having trouble answering this quadratic equation that my teacher at school used. I do not know where she had gotten this equation from.
Equation:
$$x^2-(k+1)x+k+1=0$$
Quadratic Formula:
$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$
My teacher got the final answer of $$x=\frac{(k+1)\pm k}{2}$$
I have asked my teacher on how to solve for x, however she said she won’t show the steps on how to solve this problem because it is simply too easy. When i don’t understand how she got that answer.
My working out
I have simply found my coefficients for x^2 and x, and my constant is k+1.
a= 1 
b= (-k-1) 
c= k+1 
I am not sure if c actually equals k+1
Then I put them into the quadratic formula.
$$x=\frac{-(-k-1)\pm\sqrt{(-k-1)^2-4\times 1\times k+1}}{2\times 1}$$
Which then i get
$$x=\frac{(k+1)\pm\sqrt{k^2+1-4k-3}}{2}$$
I don’t know what to do next and if my answer can be correct, please help.
 A: Apparently, the problem is not that easy after all. What your teacher believes is the solution is obviously wrong. After all, she suggests that $\frac12$ is one of the solutions, but 
$$ (\tfrac12)^2-(k+1)\tfrac12+k+1\ne 0$$
A: You forgot that $(-k-1)^2 \ne (-k)^2 + (-1)^2$. Where's the middle term?
You should have instead:
$$x=\frac{-(-k-1)\pm\sqrt{k^2+2k+1-\color{red}{4(k+1)}}}{2\times 1}$$
$$x=\frac{k+1\pm\sqrt{k^2-2k-3}}{2}$$
which does not simplify to your teacher's answer (as others have described).
A: Hint $:$ $(-k-1)^2 = (k+1)^2 =k^2 +2k+1.$
A: Your choice of parameters are correct so you should get that $$\sf{x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{(k+1)\pm\sqrt\Delta}{2}}$$ where $\sf{\Delta=(-(k+1))^2-4(1)(k+1)=k^2+2k+1-4k-4=k^2-2k-3\ne k^2}$ so your teacher is wrong nonetheless.
A: It is often a good idea to check the results by numerical examples.
If we set $k=1$, the equation becomes $x^2-x+1=0$, and your teacher's answer would be
$$x=\frac12\ !?$$
Now,
$$\frac{(k+1)\pm\sqrt{(k+1)^2-4(k+1)}}2=\frac{(k+1)\pm\sqrt{(k+1)(k-3)}}2$$
yields
$$\frac{1\pm i\sqrt3}2,$$ which works.
A simpler check is possible with $k=3$, giving $x=2$ indeed a solution of $x^2-4x+4=0$.

A check is not a proof, but it can increase your trust in a solution.
