Question

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

Answer 1

1)3х²–14х+16=0 D=(-14)²-4·3·16=196-192=4
x₁=(14-2)/6=2    или  х₂=(14+2)/6=8/3

2)4х²–36х+77=0

D=(-36)²-4·4·77=1296-1232=64
x₁=(36-8)/8=28/8=3,5    или  х₂=(36+8)/8=44/8=5,5
 
3)25х²-10х+1=0

D=(-10)²-4·25·1=0
x₁=х₂=(10-0)/50=0,2  

4) 6х(2х+1)=5х+1
12х²+6х=5х+1
12х²+х-1=0
D=1²-4·12·(-1)=49
x₁=(-1-7)/24=-1/3    или  х₂=(-1+7)/24=1/4

Answer 2

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

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