Помогите решить систему ривнянь пж.

$\left\{\begin{array}{rcl}{3y-x$=4}\\{(x^2-4xy-2y^2=3} \\\end{array}\right \\\\\\\left\{\begin{array}{rcl}{x$=3y-4}\\{(x^2-2*2xy+4y^2)-6y^2=3}\\\end{array}\right \\\\\\\left\{\begin{array}{rcl}{x$=3y-4}\\{(x-2y)^2-6y^2=3}\\\end{array}\right\\\\\\\left\{\begin{array}{rcl}{x$=3y-4}\\{(3y-4-2y)^2-6y^2=3}\\\end{array}\right\\\\\\\left\{\begin{array}{rcl}{x$=3y-4}\\{(y-4)^2-6y^2=3} \\\end{array}\right\\\\$

$\left\{\begin{array}{rcl}{x$=3y-4} \\{y^2-8y+16-6y^2=3} \\\end{array}\right\\\\\left\{\begin{array}{rcl}{x$=3y-4} \\{-5y^2-8y+13=0} \\\end{array}\right$

Решим уравнение -5y²-8y+13 = 0

5y²+8y-13 = 0

D = 64 + 4 * 5 * 13 = 64 + 260 = 324

y₁ = (-8+18)/10 = 1 ⇒ x₁ = 3*1-4 = -1

y₂ = (-8-18)/10 = -2.6 ⇒ x₂ = 3*(-2.6) - 4 = -7.8 - 4 = -11.8

Ответ: (-1, 1) и (-11.8, -2.6)