9 Класс! Помогите решить систему уравнений!!!

Решите задачу:

$\left \{ {{y = 2-3x} \atop { x^{2} -xy=3,36} \right.$
$x^{2} -x(2-3x)=3,36$
$x^{2} -2x+3 x^{2} -3,36=0$
$4 x^{2} -2x-3,36=0$
$400 x^{2} -200x-336=0$
$50 x^{2} -25x-42=0$
$D = b^{2} -4ac= (-25)^{2} -4*50*(-42)=9025$
$x_{1} = \frac{-b- \sqrt{D} }{2a} = \frac{25-95}{2*50} = \frac{-70}{100} = -0,7$
$x_{2} = \frac{-b+\sqrt{D} }{2a} = \frac{25+95}{a*50} = \frac{120}{100} = 1,2$
$y_{1} = 2-3 x_{1} = 2+3*0,7=4,1$
$y_{2} = 2-3 x_{2} =2-3*1,2=-1,6$