Odpowiedź:
[tex]f(x)=3(x+5)(x-4)\\\\f(x)=(3x+15)(x-4)\\\\f(x)=3x^2-12x+15x-60\\\\f(x)=3x^2+3x-60\ \ /:3\\\\f(x)=x^2+x-20\\\\\\\Delta=b^2-4ac=1^2-4\cdot1\cdot(-20)=1+80=81\\\\\\p=\frac{-b}{2a}=\frac{-1}{2\cdot1}=-\frac{1}{2}\\\\q=\frac{-\Delta}{4a}=\frac{-81}{4\cdot1}=\frac{-81}{4}=-20\frac{1}{4}\\\\\\f(x)=a(x-p)^2+q\\\\f(x)=1(x-(-\frac{1}{2})^2+(-20\frac{1}{4})\\\\f(x)=(x+\frac{1}{2})^2-20\frac{1}{4}[/tex]
