Решить номер 167, со всеми пояснениями

$\left \{ {(3x-y)^{2}=12-3x+y } \atop {x+y=5}} \right.\\\\\left \{ {{y=5-x} \atop {(3x-5+x)^{2}=12-3x+5-x }} \right.\\\\ \left \{ {{y=5-x} \atop {(4x-5)^{2}=17-4x }} \right\\\\\left \{ {{y=5-x} \atop {16x^{2}-40x+25-17+4x=0 }} \right.\\\\\left \{ {{y=5-x} \atop {16x^{2}-36x+8=0 }} \right.\\\\\left \{ {{y=5-x} \atop {4x^{2}-9x+2=0 }} \right.$

$4x^{2}-9x+2=0\\\\D=(-9)^{2}-4*4*2=81-32=49=7^{2}\\\\x_{1}=\frac{9+7}{8}=2\\\\x_{2}=\frac{9-7}{8}=0,25\\\\y_{1}=5-2=3\\\\y_{2} =5-0,25=4,75$

Ответ : ( 2 ; 3) , ( 0,25 ; 4 75)