Решите уравнение: 1) 5x+x-4=0 2) 20x²+x-1=0 3) 8x²+10x+3=0 4)-6x²+7-2=0

1) 5х²+х-4=0
D=1²-4·5·(-4)=81
$x_{1,2}$ = $\frac{-1+-9}{10}$
$x_{1}$ = $\frac{-1+9}{10}$ = $\frac{8}{10}$ =0,8
$x_{2}$ = $\frac{-1-9}{10}$ =- $\frac{10}{10}$ =-1
2) $20x^{2}$ +х-1=0
D=1²-4·20-1=81
$x_{1, 2}$ = $\ \frac{-1+-9}{40}$
   $x_{1}$ = $\frac{-1+9}{40}$ = $\frac{8}{40}$ = $\frac{1}{5}$
   $x_{2}$ = $\frac{-1-9}{40}$ = $\frac{-10}{40}$ =- $\frac{1}{4}$
  3) $8x^{2}$ +10х+3=0
D=10²-4·8·3=4
$x_{1,2}$ = $\frac{-10+-2}{16}$
$x_{1}$ = $\frac{-10+2}{16}$ =- $\frac{8}{16}$ =-0,5
$x_{2}$ = $\frac{-10-2}{16}$ =- $\frac{12}{16}$ =- $\frac{3}{4}$ =-0,75
4)-6х²+7х-2=0 | ÷(-1)
6х²-7х+2=0
D=(-7)²-4·6·2=49-48=1
$x_{1,2}$ = $\frac{7+-1}{12}$
$x_{1}$ = $\frac{7-1}{12}$ = $\frac{6}{12}$ =0,5
$x_{2}$ = $\frac{7+1}{12}$ = $\frac{8}{12}$ = $\frac{2}{3}$