Find the range of values of $k$ for which the following equation has real roots. Find the range of values of $k$ for which the following equation $x^2+(1-k)x=k$  has real roots.
I tried it, for real roots, $b^2-4ac \geqslant  0$
$x^2+(1-k)x-k=0$
$(1-k)^2-4*1*(-k)\geqslant0$
${(1)^2-2\cdot1\cdot k+(k)^2}-4(-k)\geqslant0$
$1-2k+k^2+4k\geqslant0$
$k^2+2k+1\geqslant0$
$k^2+k+k+1\geqslant0$
$(k+1)(k+1)\geqslant0$
Solving the equation I get $k\geqslant -1$. Is this right?
In the book, the answer given is "all values of $k$"
What does that mean?
 A: Recall the formula $x_{1,2}=\frac{-b+-\sqrt{b^2-4ac}}{2a}$. This means that $x_{1,2}$ has real values if and only if $b^2-4ac$ is positive (because it is under square root). Now, since you've found that $b^2-4ac=(1+k)(1+k)=(1+k)^2\geqslant0$ it means that for any $k\in\mathbb{R}$ we can find $x_{1,2}$, so it has real roots for any $k\in\mathbb{R}$.
A: The discriminant ($\Delta$) of the quadratic equation $x^2 + (1-k)x - k = 0$ is (using the $b^2 - 4ac$ "mnemonic") : 
$$\begin{align}\Delta &= (1-k)^2 -4(1)(-k)\\
&=1 - 2k + k^2 + 4k\\
&= 1 + 2k + k^2\\
&= (k + 1)^2\end{align}$$
Now, a quadratic equation has only real roots if and only if $\Delta \ge 0$. This condition is met for all $k$, provided that $k \in \mathbb{R}$ (because the square of any real number is at least $0$).
It will generally fail otherwise (take $k = i = \sqrt{-1}$, the imaginary unit for example). It does not work for all numbers in general, if we were to consider complex and hypercomplex numbers.
So the required values of $k$ is $k\in\mathbb{R}$. 
