Quartic with $4 $ equidistant roots Today I got the problem $(x^2 -1)(x^2 -4)=k$, and I have no idea how to prove this algebraically. $K$ is a real, non-zero number that makes the equation have $4$ distinct real equidistant roots. Solve for $k$ algebraically.

What I have attempted:
I have tried solving for x and then in turn attempting to solve for k
$$(x^2 -1)(x^2 -4)=k$$
$$ (x^4 -4x^2-1x^2+4) = k$$
$$ x^4 - 5x^2 + 4 - k = 0 $$
$$ x^2 = {-b\pm\sqrt{b^2-4ac} \over 2a} $$  
$$ x^2 = {5\pm\sqrt{25-4(4-k)} \over 2} $$
$$ x^2 = {5\pm\sqrt{9+4k} \over 2} $$
$$ x^2 = {5+\sqrt{9+4k} \over 2} $$ and $$ x^2 = {5-\sqrt{9+4k} \over 2} $$
Therefore 
$$ x_1 = \sqrt {5+\sqrt{9+4k} \over 2} $$ 
$$ x_2 = -\sqrt {5+\sqrt{9+4k} \over 2} $$ 
$$ x_3 = \sqrt  {5-\sqrt{9+4k} \over 2}  $$ 
$$ x_4 = -\sqrt  {5-\sqrt{9+4k} \over 2}  $$ 
Now I am stuck trying to solve for k.. The answer is $k$ = 7/4 but having trouble solving for it algebraically..
 A: As the roots are equidistant, let the roots be  $a\pm d,a\pm3d$
using Vieta's formula, immediately we have $$a+d+a-d+a+3d+a-3d=0\iff a=0$$
and $5=(-d)(d)+(-d)(3d)+(-d)(-3d)+d(3d)+(d)(-3d)+(3d)(-3d)$
Now $4-k= d\cdot-d\cdot3d\cdot-3d$
Can you take it from here?
In fact $$x^4-5x^2+4-k=(x^2-d^2)(x^2-9d^2)=x^4-10d^2x^2+9d^4$$
$\implies10d^2=5$ and $4-k=9d^4$
A: You have found $x^2=A+B$ and $x^2=A-B$ with $$A={5\over2},\qquad B={1\over2}\sqrt{9+4k}\ .\tag{1}$$ It follows that the four roots are (in increasing order):
$$x_1=-\sqrt{A+B},\quad x_2=-\sqrt{A-B},\quad x_3=\sqrt{A-B},\quad x_4=\sqrt{A+B}\ .$$
The condition $x_3-x_2=x_4-x_3$ leads to
$$\sqrt{A+B}=3\sqrt{A-B}\ ,$$
and this in turn implies $10B=8A$. Plugging in $(1)$ and solving for $k$ leads to $k={7\over4}$, so that $A+B={9\over2}$, $A-B={1\over2}$. This then leads to
$$x_1=-{3\over2}\sqrt{2},\quad x_2=-{1\over2}\sqrt{2},\quad x_3={1\over2}\sqrt{2},\quad x_4={3\over2}\sqrt{2}\ .$$
One easily checks that these findings fulfill all requirements.
