What's special about seven?

You must see the excellent points of @Joel Tan and @Abhinav Raichur below my solution.

Let $P(x)=(x^2+a)^2+a-x \\ P(x)=x^4+2x^2a-x+a(a+1)$ Leading and constant terms of polynomial gives us an idea for writing polynomial as product of two quadratics: $P(x)=x^4+2x^2a-x+a(a+1)=(x^2+bx+a)(x^2+cx+a+1)$ There is no $x^3$ terms in polynomial, It follows that $c=-d$ and using the coefficient of $x$ in both sides gives $cax-c(a+1)x=-x \\ -cx=-x \\ c=1$ Therefore we have $P(x)=(x^2-x+a)(x^2+x+a+1)$ One can simplify this to see that the factoring is right. Now, let $P_1(x)=x^2-x+a$ and $P_2(x)=x^2+x+a+1$ . Equation has four distinct real roots, so discriminant of both of the quadratics must be positive, which means $\Delta_1=1-4a>0 \\ a<\frac{1}{4} \\ \Delta_2=-4a-3>0 \\ a<-\frac{3}{4}$ Hence we must have $a<-\frac{3}{4}$ . It can be seen that for $a<-\frac{3}{4}$ two polynomials does not share common roots (how?). Now, the roots will be $\frac{1 \pm \sqrt{1-4a}}{2}$ for $P_1(x)$ and $\frac{-1 \pm \sqrt{-4a-3}}{2}$ for $P_2(x)$ , since $1-4a>-4a-3$ roots of $P_1(x)$ are greater than $P_2(x)$ , so greater root of $P_1(x)$ which is $x_4=\frac{1 + \sqrt{1-4a}}{2}$ is the largest root of $P(x)$ and the smaller root of $P_2(x)$ which is $x_1=\frac{-1 - \sqrt{-4a-3}}{2}$ is the smallest root of $P(x)$ , finally we must have $x_4-x_1=\frac{1 + \sqrt{1-4a}}{2}-\frac{-1 - \sqrt{-4a-3}}{2}=7 \\ \sqrt{1-4a}+\sqrt{-4a-3}=12 \\ 1-4a-4a-3+2\sqrt{(1-4a)(-4a-3)}=144 \\ \sqrt{(1-4a)(-4a-3)}=73+4a \\ (-1-4a)(-4a-3)=16a^2+584a+73^2 \\ 16a^2+8a-3=16a^2+584a+73^2 \\ a=-\frac{73^2+3}{584-8}=-\frac{1333}{144}$