Question

помогите решить
(2у - 3у²) =(2у) ²-2 ×2у×3у+(3у)²=​

Answer 1

(2у - 3у²) = (2у) ² - 2 × 2у × 3у + (3у)²

2y - 3y² = 4у² - 12у² + 9у²

2y - 3y² = у²

-3у² - у² = 2y

-4у² = 2y

-4у² - 2y = 0

-2у (2y - 1) = 0

y (2y - 1) = 0

y = 0

или

2у - 1 = 0

2у = 1

у = 1/2