Question

(y-3)²-4(3-y)=0
решить уравнение ​

Answer 1

Ответ:

${y}^{2} - 6y + 9 - 12 + 4y = 0 \\ {y}^{2} - 2y - 3 = 0 \\ d = 4 + 12 = 16 = {4}^{2}$

$y1 = \frac{6 + 4}{2} = \frac{10}{2} = 5 \\ y2 = \frac{6 - 4}{2} = \frac{2}{2} = 1$