I'm sure there's an easier way of solving it than the way I did, but I'm not sure what it could be. Never dealt with a problem like this before.
Anyway, I just plugged in and tested. Chose random values for a, b, c, and d, which follow the rule 0 < a < b < c < d:
a = 1
b = 2
c = 3
d = 4
[tex]\sf ax^2+(1-a(b+c))x+abc-d)[/tex]
[tex]\sf 1x^2+(1-1(2+3))x+(1)(2)(3)-(4))[/tex]
Simplify into standard form:
[tex]\sf x^2+(1-1(5))x+6-4[/tex]
[tex]\sf x^2+(1-5)x+2[/tex]
[tex]\sf x^2-4x+2[/tex]
Use the quadratic formula to solve:
[tex]\sf x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]
For functions in the form of [tex]\sf ax^2+bx+c[/tex]. So in this case:
a = 1
b = -4
c = 2
Plug them in:
[tex]\sf x=\dfrac{4\pm\sqrt{(-4)^2-4(1)(2)}}{2(1)}[/tex]
Solve for 'x':
[tex]\sf x=\dfrac{4\pm\sqrt{16-8}}{2}[/tex]
[tex]\sf x=\dfrac{4\pm\sqrt{8}}{2}[/tex]
[tex]\sf x\approx\dfrac{4\pm 2.83}{2}[/tex]
[tex]\sf x\approx 0.59,3.41[/tex]
So the answer would be A.
