Решите систему уравнений х-2у=5,ху=12

$\left \{ {{x-2y=5} \atop {xy=12}} \right. \left \{ {{x=5+2y} \atop {xy=12}} \right.$

$y(5+2y)=12$
$5y+2 y^{2} =12$
$2 y^{2} +5y-12=0$
$a=2;b=5;c=-12$
$D= b^{2} -4ac$
$D= 5^{2} -4*2*(-12)=25+96=121$
$\sqrt{D} =11$
$y= \frac{-b+-\sqrt{D}}{2a}$
$y_{1} = \frac{-5+11}{2*2} = \frac{6}{4} = \frac{3}{2} =1 \frac{1}{2} =1,5$
$y_{2} = \frac{-5-11}{2*2} = \frac{-16}{4} = -\frac{4}{1} =-4$

$x_{1} =5+2*1,5=5+3=8$
$x_{2} =5+2*(-4)=5-8=-3$

Ответ: $(8;1,5);(-3;-4)$