Question

2) (1 - y) + 8y = 7 + y(3 - y);

Answer 1

Ответ:

Объяснение:

(1 - y) + 8y = 7 + y(3 - y)

1 - y + 8y = 7 + 3y - y²

1 - y + 8y - 7 - 3y + y² = 0

$\[\begin{gathered}{y^2}+4y-6=0\hfill\\D={b^2}-4ac={4^2}-4\cdot 1\cdot (-6)=16+24=40\hfill\\{y_{1;2}}=\frac{{-b\pm\sqrt D}}{{2a}}=\frac{{-4\pm\sqrt {40}}}{{2\cdot1}}=\frac{{-4\pm2\sqrt {10}}}{2}=\frac{{2(-2\pm\sqrt {10})}}{2}=-2\pm\sqrt {10}\hfill\\{y_1}=-2+\sqrt {10}\hfill\\{y_2}=-2-\sqrt {10}\hfill\\\end{gathered}\]$