Question

задание на фото, помогите пожалуйста

Answer 1

$\left \{ {{y=x^2-4} \atop {y=2y-7}} \right.\\x=2(x^2-4)-7.\\x=2x^2-8-7.\\x-2x^2+15=0.\\2x^2-x-15=0.\\D = b^2 - 4ac = (-1)^2 - 4*2*(-15) = 1 + 120 = 121.\\\sqrt{D} =11.\\x_{1} =\frac{-10}{4}=-\frac{5}{2} .\\x_{2} =\frac{12}{4}=3.\\y_{1} =(-\frac{5}{2})^2-4=\frac{9}{4}.\\y_{2} =3^2-4=5.\\\\(x_{1} ,y_{1} )=(-\frac{5}{2},\frac{9}{4}) .\\(x_{2} ,y_{2} )=(3,5) .$