Yes. it seems a difficult problem. Applying the condition is tricky...

Let us split this into multiple sum..

because you have constraint( B2-4AC >=0 ) here and lower and upper bounds ( 0 and 1). So this would make multiple cases.

*case1 2sqrt(AC) <=1 and 4C <=1*
the above condition we can include in the limit as follows

C ranges 0 to 1/4

A ranges 0 to min(1/(4C),1)

**=> 0 to 1**
and B ranges 0 to 2sqrt(AC)

So the area is = int(C) int(A) int(B,0 to 2sqrt(AC)) fC fA fC

= int(C) int(A , 0 to 1 )

**2sqrt(AC) ** fA fC

= int(C, 0 to 1) {

**2sqrt(AC) A *2/3**}at A = 1 fC

= int(C, 0 to 1) {

**sqrt(C) *2/3**}fC

= {2/3

**sqrt(C)C *2/3 }**at C=1/4

= 1/18

*case2 2sqrt(AC) <=1 and 4C >=1*
the above condition we can include in the limit as follows

C ranges 1/4 to 1

A ranges 0 to 1/(4C)

and B ranges 0 to 2sqrt(AC)

So the area is = int(C) int(A) int(B,0 to 2sqrt(AC)) fC fA fC

= int(C) int(A , 0 to 1 )

**2sqrt(AC) ** fA fC

= int(C, 0 to 1) {

**2sqrt(AC) A *2/3**}at A = 1/4C fC

= int(C, 0 to 1) {

**1/4C *2/3**}fC

= {

**1/6 log(C) }**at C=1 - {

**1/6 log(C) }**at C=1/4

*case3 2sqrt(AC) >=1 *and C<=1/4

the above condition we can include in the limit as follows

C ranges 0 to 1/4

A ranges 1/4C to 1

**=> 1 to 1**
and B ranges 0 to 2sqrt(AC)

**area =0**
*case4 2sqrt(AC) >=1 and *C >=1/4

the above condition we can include in the limit as follows

C ranges 1/4 to 1

A ranges 1/(4C) to 1

and B ranges 0 to 1

calculate area..

You can sum the four cases and get the proabability.

Please cross check the integral, because i haven't done integration for long time

please let me know if you want to have clarification on any steps apart from integration.