I want everything real

The given equation $x^{4}-(a-3)(a-2)x^{2}-(a-1)(a-2)=0 \rightarrow (1)$

can be written as: $t^{2}-(a-3)(a-2)t-(a-1)(a-2)=0$

where $t=x^{2}$

Now, the roots of this equation have to be positive or else the roots of $(1)$ will be imaginary, thus

$i)$ The product of the roots $-(a-1)(a-2) \geq 0$

$ii)$ The sum of the roots $(a-3)(a-2) \geq 0$

From $(i)$ we have $1 \leq a \leq 2$

and from $(ii)$ we have $a \leq 2, 3 \leq a$

The overlapping interval is $1 \leq a \leq 2$

Hence the sum of the integral values of $a = 1+2 = \boxed{3}$

first let $t=x^2$ 2 $t^2-(a^2 -5a+6)t-(a^2-3a+2)=0$ by the quadratic formula, $t=\dfrac{a^2-5a+6\pm\sqrt{(a^2-5a+6)^2+4(a^2-3a+2)}}{2}$ hence $x=\sqrt{\dfrac{a^2-5a+6\pm\sqrt{(a^2-5a+6)^2+4(a^2-3a+2)}}{2}},-\sqrt{\dfrac{a^2-5a+6\pm\sqrt{(a^2-5a+6)^2+4(a^2-3a+2)}}{2}}$ for all x to be real , $\dfrac{a^2-5a+6-\sqrt{(a^2-5a+6)^2+4(a^2-3a+2)}}{2}\geq 0$ now ,we get $-\sqrt{(a^2-5a+6)^2+4(a^2-3a+2)}\geq a^2-5a+6$ $(a^2-5a+6)^2+4(a^2-3a+2)\leq (a^2-5a+6)^2$ $a^2-3a+2\leq 0$ $(a-1)(a-2)\leq 0$ $1\leq a\leq 2$ hence, the sum of the integral values are $1+2=\boxed{3}$

$The~given ~equation~is~quadratic~in~x^2.~So~real~values~~x^2~can~ not~be~negative. \\ (a^2-5a+6)=(a-2)(a-3),~~~~~~(a^2-3a+2)=(a-2)(a-1).\\ So ~the~product~of~roots~is~~-~ (a-2)(a-1), ~which ~is~negative~or~0.\\ So~only~possible~solution~is~if~x^2=0.\\ So ~if~a=2,~it~is~easy~to~see~x^4=0.\ ~only~roots~are~real~0s.\\ If~a=1,~~x^4-2x^2-0=0,~gives~x^2=0~~or~~x^2=2.~~ Solution~are~~0,~0,~-~\sqrt2,~\sqrt2~real~roots.\\ So~a=1~or~a=2~are~only~solutions.\\ So ~1+2=\Large \color{#D61F06}{3}.$