Решите систему управления х+у=5 ху=4

$\left \{ {{x+y=5} \atop {xy=4}} \right.$
$\left \{ {{x=5-y} \atop {xy=4}} \right.$
(5-y)y=4
5y- $y^{2}$ = 4
- $y^{2}$ + 5y - 4 = 0
a =-1 b=5 c=-4
D = $b^{2}$ -4ac
D = $5^{2}$ - 4 * (-1) * (-4) = 25 - 16 = 9
$\sqrt{D} = \sqrt{9} = 3$
$y_{1} = \frac{-b+ \sqrt{D} }{2a} = \frac{-5+3}{-2} = \frac{-2}{-2} = 1$
$y_{2} = \frac{-b- \sqrt{D} }{2a} = \frac{-5-3}{-2} = \frac{-8}{-2} =4$
$\left \{ {{y_{1} =1 ; y_{2}=4 } \atop {x=5-y}} \right.$
$x_{1} = 5 - y_{1} = 5-1=4$
$x_{2} = 5- y_{2} = 5-4=1$
Ответ. (4;1), (1;4)