Помогите решить

2x в квадрате + ху = 40

3 х-у=10

{ 2x² + xy = 40

{ 3x - y = 10

{ y = 3x - 10

{ 2x² + x (3x - 10) = 40

5x² - 10x - 40 = 0

x² - 2x - 8 = 0 D = b²-4ac = 4+32 = 36

x₁ = (-b+√D) / 2a = (2+6) : 2 = 4 y₁ = 3x - 10 = 2

x₂ = (-b-√D) / 2a = (2-6) : 2 = - 2 y₂ = 3x - 10 = - 16

Ответ: {4; 2}; {-2; - 16}

Y=3x-10

2^2+x (3x-10) = 40

2x^2+3x^2-10x-40=0

5x^2-10x-40=0

D=100+800=900

x1=10-30/10=-2

x2=10+30/10=4

x1=-2

y1=3 * (-2) - 10=-16

x2=4

y2=3*4-10=2