((Решите уравнения.))

$a)3 {x}^{2} - 5x + 2 = 0 \\ a = 3 \\ b = - 5 \\ c = 2$

$d = {b}^{2} - 4ac = ({ - 5})^{2} - 4 \times 3 \times 2 = \\ = 25 - 24 = 1$

$x1 = \frac{ - 5 + 1}{6} = - \frac{4}{6} = - \frac{2}{3}$

$x2 = \frac{ - 5 - 1}{6} = - \frac{6}{6} = - 1$

Б)

${x}^{2} + 3x + 3 = 0 \\ d = 9 - 12 = - 3$

Нет решения

В)

$5 {y}^{2} - 4y = 1 \\ y(5y - 4) = 1 \\ y = 1$

$5y - 4 = 1 \\ 5y = 1 + 4 \\ 5y = 5 \\ y = 5 \div 5 \\ y = 1$