Найдите корни уравнения: а) (x+3) (x-4) = -12 б) 1 2/3 t + (2t + 1) (1/3 t - 1 ) = 0 в) 3x (2x + 3) = 2x (x + 4, 5) + 2 г) (x -1) (x+1) = 2(x^2 - 3) распишите поподробней .

X^2-4x+3x-12=-12
x^2-4x=0
x(x-4)=0
x=0, x=4

Решение
а) (x+3) (x-4) = -12
x² - x - 12 = - 12
x² - x = 0
x(x - 1) = 0
x₁ = 0
x - 1 = 0
x₂ = 1

б) 1 2/3 t + (2t + 1) (1/3 t - 1 ) = 0
5/3 t + (2/3)*t² - 2t + (1/3)t - 1 = 0
(2/3)*t² - 1 = 0
2t² - 3 = 0
t² = 3/2
t₁ = - √(3/2)
t₁ = - √6/2
t₂ = √6/2

в) 3x (2x + 3) = 2x (x + 4, 5) + 2
6x² + 9x = 2x² + 9x + 2
4x² = 2
x² = 1/2
x₁ = - 1/√2
x₂ = 1/√2

г) (x -1) (x+1) = 2(x^2 - 3)
x² - 1 = 2x² - 6
x² = 5
x₁ = - √5
x₂ = √5