wyznaczam dziedzinę:
4x^2+4x+1\ne0
(2x+1)^2\ne0
2x+1\ne0 (także II mianownik)
2x\ne-1
x\ne-\frac{1}{2} , D=R \ {-1/2)
\frac{3-x}{4x^2+4x+1} - \frac{5x}{2x+1} = 0
\frac{3-x}{4x^2+4x+1} = \frac{5x}{2x+1}
\frac{3-x}{(2x+1)^2}-\frac{5x}{2x+1}=0
\frac{3-x-5x(2x+1)}{(2x+1)^2}=0
\frac{3-x-10x^2-5x}{(2x+1)^2}=0
10x^2-6x+3=0
a=-10, b=-6, c=3
\Delta=b^2-4ac=36-4*(-10)*3=36+120=156
\sqrt\Delta=\sqrt{156}=\sqrt{4*39}=2\sqrt{39}
x_1=\frac{-b-\sqrt\Delta}{2a}=\frac{6-2\sqrt{39}}{2*(-10)}=\frac{-2(\sqrt{39}-3)}{-20}=\frac{\sqrt{39}-3}{10}
x_2=\frac{-b+\sqrt\Delta}{2a}=\frac{6+\sqrt{39}}{-20}=\frac{-2(-3-\sqrt{39})}{-20}=\frac{-3-\sqrt{39}}{10}