Calculating the percentage of (+)-2- butanol in the sample

Specific rotation of (+)-2-butanol is 12.5$$^{\circ}$$ . A sample of 2-butanol containing both the enantiomers was found to have specific rotation value of –2.8$$^{\circ}$$ . What is the percentage of (+)-2- butanol in the sample?

I attempted it by letting the percentage of (+)-2-butanol be $$x \%$$ and therefore percentage of enantiomers in the mixture would be $$2x \%$$ due to equal contribution by (-)-2-butanol. Since the net specific rotation of the mixture is given to be negative, we know that percentage of excess of (-)-2-butanol would be $$(100-2x)\%$$. It is also given that specific rotation of pure (+)-2-butanol is 12.5$$^{\circ}$$, but I am unable to understand how is that useful here.

In it's solution, they have calculated the percentage of (-)-2-butanol as $$\dfrac{2.8}{12.5} \times 100=22.4 \%$$ and thus percentage of (+)-2-butanol came out to be $$\dfrac{100-22.4}{2}=38.8\%$$.

I didn't understand this last step, why is the division of 2.8 and 12.5 giving the percentage of (-)-2-butanol?

Since the optical rotation of (+)-2-butanol is 12.5 degrees, the optical rotation of (-)-2-butanol is -12.5 degrees. If the mole fraction of the (+) isomer is $$x$$ and the mole fraction of the (-) isomer is $$1-x$$, then $$-2.8 = 12.5x -12.5(1-x)$$
Solving this simple linear equation gives $$x = 0.388$$, which when converted to mole percentage, is $$38.8\%$$.