So if the quantities are x and y, then xy=3.75 and x+y=1, so y=1-x, and we can substitute for y in the other equation:
x(1-x)=3.75, x-x2=3.75, x2-x+3.75=0 (same as quadratic 4x2-4x+15=0),
x2-x+0.25-0.25+3.75=0 (completing the square),
(x-0.5)2+3.50=0,
(x-0.5)2=-3.50. The solution is complex, involving imaginary i=√-1.
x-0.5=±i√3.50, x=0.5±i√3.5.
I suspect that xy=-3.75 rather than xy=3.75 so this simplifies the solution so that there is no complex solution:
x2-x+0.25-0.25-3.75=0 (completing the square),
(x-0.5)2-4=0,
(x-0.5)2=4,
x-0.5=2 or -2, so x=2.5 or -1.5. Then y=-1.5 or 2.5.
With the correction the quadratic becomes 4x2-4x-15=0=(2x+3)(2x-5).