Question

найдите корень уравнения: 3(x-2)+(x-4)=0
(у-3)^2-4(3-у)=0

Answer 1

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