Помогите пожалуйста 2 задание сделать

Решите задачу:

$- 0.2 {x}^{2} + 4 = 0 \\ - 0.2 {x}^{2} = - 4 \\ {x}^{2} = - 4 \div ( - 0.2) \\ {x }^{2} = 20 \\ x1 = - 2 \sqrt{2} \\ x2 = 2 \sqrt{2}$
$\frac{ {x}^{2} }{3} + \frac{x}{9} = 0 \\ \frac{1}{3} x(x + \frac{1}{3} ) = 0 \\ x = 0 \\ x = - \frac{1}{3}$
${(2x - 1)}^{2} = 1 - 4x \\ 4 {x}^{2} - 4x + 1 = 1 - 4x \\ 4 {x}^{2} = 0 \\ x = 0$
$3 - (4x + 1)(3 - x) = x \\ 3 - 12x + 4 {x}^{2} - 3 + x = x \\ 4 {x}^{2} - 12x = 0 \\ 4x(x - 3) = 0 \\ x = 0 \\ x = 3$
$3 - 0.4 {x}^{2} = 0 \\ 0.4 {x}^{2} = 3 \\ {x}^{2} = 7.5 \\ x = - \frac{ \sqrt{30} }{2} \\ x = \frac{ \sqrt{30} }{2}$
$\frac{ {x}^{2} }{4} - \frac{x}{2} = 0 \\ \frac{x}{2} ( \frac{x}{2} - 1) = 0 \\ x = 0 \\ x = 2$
${(3x + 2)}^{2} = 4 + 12x \\ 9 {x}^{2} + 12x + 4 = 4 + 12x \\ 9 {x}^{2} = 0 \\ x = 0$
${x}^{2} - (2x - 3)(1 - x) = 0 \\ {x }^{2} - 2x + 2 {x}^{2} + 3 - 3x = 0 \\ 3 {x}^{2} - 5x + 3 = 0 \\ net \: otweta$