Пожалуйста, помогите Пятое и шестое, отмеченное красными точками

[latex]3x^2-5x-2=0\\ D=25+24=49; \sqrt{D}=7\\\\ x_{1/2}= \frac{5\pm7}{6}\\\\x_1=- \frac{2}{6} = \frac{1}{3}\\\\ x_2= \frac{12}{6}\\\\ 3x^2-5x-2=(x-2)(3x-1) [/latex] [latex] \frac{-5a^2+4a+1}{1-25a^2}=(*)= \frac{(-1-5a)(a-1)}{(1-5a)(a-1)}=- \frac{(1+5a)(a-1)}{(1+5a)(1-5a)}=- \frac{a-1}{1-5a}= \frac{a-1}{5a-1} \\\\ (*)-5a^2+4a+1=0\\ D=16+20=36; \sqrt{D}=6\\\\ x_{1/2}= \frac{-4\pm6}{-10} \\\\ x_1=1\\ x_2=- \frac{1}{5} \\\\[/latex]

5. 1) [latex]3x^2-5x-2=0[/latex] [latex]D=25-4*3*(-2)=25+24=49[/latex] [latex]x_{1}= \frac{5-7}{6}=- \frac{2}{6}=- \frac{1}{3} [/latex] [latex]x_{2}= \frac{5+7}{6}=2 [/latex] [latex]3x^2-5x-2=3(x+ \frac{1}{3} )(x-2)=(3x+1)(x-2)[/latex] 6. 1) [latex]-5a^2+4a+1=0[/latex] [latex]D=16-4*(-5)=36[/latex] [latex]a_{1}= \frac{-4-6}{-10}=1 [/latex] [latex]a_{2}= \frac{-4+6}{-10}=-0.2 [/latex] [latex]-5a^2+4a+1=-5(a-1)(a+0.2)=(1-a)(5a+1)[/latex] [latex] \frac{-5a^2+4a+1}{1-25a^2}= \frac{(1-a)(5a+1)}{(1-5a)(1+5a)}= \frac{1-a}{1-5a} [/latex]