[latex]a) 3(x-sqrt3)^{2}-5(x-sqrt3)+2=0\\t=x-sqrt3\\3t^2-5t+2=0\\Delta_t=(-5)^2-4cdot3cdot2=25-24=1; sqrtDelta=sqrt1=1\\t_1=frac{5-1}{2cdot3}=frac{2}{3}; t_2=frac{5+1}{2cdot3}=1\\x_1-sqrt3=frac{2}{3} vee x_2-sqrt3=1\\x_1=frac{2}{3}+sqrt3 vee x_2=1+sqrt3[/latex] [latex]b) 2(x+frac{3}{7})+3(x+frac{3}{7})-2=0\\t=x+frac{3}{7}\\2t+3t-2=0\\5t=2 /:5\\t=frac{2}{5}\\x+frac{3}{7}=frac{2}{5}\\x=frac{14}{35}-frac{15}{35}\\x=frac{1}{35}\\\©DRK[/latex] [latex]Jakby w b) mialo byc rownanie kwadratowe:\\2(x+frac{3}{7})^2+3(x+frac{3}{7})-2=0\\t=x+frac{3}{7}\\2t^2+3t-2=0\\Delta=3^2-4cdot2cdot(-2)=9+16=25; sqrtDelta=sqrt{25}=5\\t_1=frac{-3-5}{2cdot2}=-2; t_2=frac{-3+5}{2cdot2}=frac{1}{2}\\x_1+frac{3}{7}=-2 vee x_2+frac{3}{7}=frac{1}{2}\\x_1=-2-frac{3}{7} vee x_2=frac{7}{14}-frac{6}{14}\\x_1=-2frac{3}{7} vee x_2=frac{1}{14}[/latex]
