Without promo, x donut = $15
$$For\ dozen\ donut\ =\frac{12\cdot15}{x}$$
$$During\ promo\ customers\ get\ 3\ free\ donuts\ for\ the\ price\ of\ x$$
$$\cos t\ of\ dozen\ donuts\ during\ promotion\ =>$$
$$\frac{12\cdot15}{x+3}$$
$$if\ donut\ price\ per\ dozen\ during\ pomo\ is\$ 2\ less\ than\ normal\ price\ per\ dozen$$ $$\frac{12\cdot15}{x+3}=\left(\frac{12\cdot15}{x}\right)-2$$
$$solve\ for\ x$$
$$\frac{180}{x+3}=\frac{180}{x}-\frac{2}{1}$$
$$\frac{180}{x+3}=\frac{180-2x}{x}$$
$$180x=\left(x+3\right)\left(180-2x\right)$$
$$180x=180x-2x^2+540-6x$$
$$2x^2+6x-540=0$$
$$2x^2+6x-540=0$$
$$\frac{2x^2}{2}+\frac{6x}{2}=\frac{540}{2}$$
$$x^2+3x=270$$
$$x^2+3x-270=0$$
$$x^2-15x+18x-270=0$$
$$\left(x^2-15x\right)+\left(18x-270\right)=0$$
$$x\left(x-15\right)+18\left(x-15\right)=0$$
$$\left(x+18\right)\left(x-15\right)=0$$
$$x+18=0\ or\ x-15=0$$
$$x=-18\ or\ x=15$$
$$Since\ the\ amount\ of\ donut\ cannot\ be\ negative,\ definitely\ x=15$$
$$Answer\ =\ A$$