Помогите 3х2–14х+16=0 4х2–36х+77=0 25х2=10х–1 6х(2х+1)=5х+1

Решите задачу:

$1)3 x^{2} -14x+16=0\\D=196-192=4 \\ \sqrt{D} =2 \\ x_{1} = \frac{14+2}{6} = \frac{16}{6} = \frac{8}{3} =2 \frac{2}{3} \\ x_{2} = \frac{14-2}{6} = \frac{12}{6} =2 \\ \\ 2)4 x^{2} -36x+77=0 \\ D=1296-1232=64 \\ \sqrt{D} =8 \\ x_{1} = \frac{36+8}{8} = \frac{44}{8} = \frac{11}{2} =5.5 \\ x_{2} = \frac{36-8}{8} = \frac{28}{8} = \frac{7}{2} =3.5$

$3) 25 x^{2} -10x+1=0\\D=100-100=0 \\ x_{0} = \frac{10}{50} =0.2 \\ \\ 4)12 x^{2} +6x-5x-1=0 \\ 12 x^{2} +x-1=0 \\ D=1+48=49 \\ \sqrt{D} =7 \\ x_{1} = \frac{-1+7}{24} = \frac{6}{24} = \frac{1}{4} =0.25 \\ x_{2} = \frac{-1-7}{24} =- \frac{8}{24} =- \frac{1}{3}$