Выполните действия: а) (2х^2-3х)-(5х-2х^2), б) -3х(2х-1), в) (3-у^2)(у-4)

Решите задачу:

$(2x^2 - 3x) - (5x - 2x^2) = 0$
$4x^2 - 8x = 0 4x(x - 2) = 0 x(x - 2) = 0 x - 2 = 0 ; x = 0 x1 = 2 ; x2 = 0$

$-3x(2x - 1) = 0 x(2x - 1) = 0 x = 0 ; 2x - 1 = 0 x1 = 0 ; x2 = 0,5$

$(3 - y^2)(y - 4) = 0 y - 4 = 0 ; 3 - y^2 = 0 y = 4 ; -y^2 = -3 y = 4 ; y^2 = 3 y1 = 4 ; y2 = \sqrt{3} ; y3 = - \sqrt{3}$