Решите систему уравнений!!!! Срочно

$\left \{ {{ log_{2} x+ log_{2} y=2} \atop {x-4y=15}} \right.$
$\left \{ {{ log_{2} x*y=2log_{2}2} \atop {x=15+4y}} \right.$
$\left \{ {{ xy=4} \atop {x=15+4y}} \right.$
$\left \{ {{ (15+4y)y=4} \atop {x=15+4y}} \right.$
$\left \{ {{ 15y+4y^2-4=0} \atop {x=15+4y}} \right.$
$4y^2+15y-4=0$
$D=225-4*4*(-4)=225+64=289$
$y_{1} = \frac{-15+17}{8} = \frac{1}{4}$
$y_{2} = \frac{-15-17}{8} = -4$

$\left \{ {{y= \frac{1}{4} } \atop {x=15+4y}} \right.$
$\left \{ {{y= \frac{1}{4} } \atop {x=15+4* \frac{1}{4} }} \right.$
$\left \{ {{y= \frac{1}{4} } \atop {x=15+1}} \right.$
$\left \{ {{y= \frac{1}{4} } \atop {x=16}} \right.$

$\left \{ {{y= -4} \atop {x=15+4y}} \right.$
$\left \{ {{y= -4} \atop {x=15+4*(-4)}} \right.$
$\left \{ {{y= -4} \atop {x=15-16}} \right.$
$\left \{ {{y= -4} \atop {x=-1}} \right.$
ответ: (16;1/4);(-4;1)