Найдите корень уравнения: 3 (x-2) + (x-4) = 0

(у-3) ^2-4 (3-у) = 0

3 (x-2) + (x-4) = 0

3x-6+x-4=0

4x-6-4=0

4x-10=0

4x=10

4x/4 = 10/4

x=10/4

x = 5/2

(y-3) ^2-4 (3-y) = 0

(y-3) ^2-12+4y=0

(y-3) (y-3) - 12+4y=0

y^2-6y+9-12+4y=0

y^2-2y+9-12=0

y^2-2y-3=0

D = (-2) ^2-4*1 * (-3) = 16

x1=-1

x2=3 = > y = 3; -1