Question

Помогите решить уравнение 6+y(y-4)=3y

Answer 1

6 + у^2 - 4y = 3y
6 + y ^2 - 4y - 3y = 0
y^2 - 7y + 6 = 0
D = 49 - 4 * 6 = 49 -24 = 25    YD = 5
x1 = (7 + 5) / 2 = 6
x2 = (7 - 5) / 2 = 1
x1 = 6
x2 = 1

Answer 2

6+y(y-4)=3y
6+y²-4y-3y=0
y²-7y+6=0

$D=7^{2}-4*6*1=49-24=25=5^{2} \ x_{1}= frac{-(-7)- sqrt{25} }{2*1} = frac{7-5}{2} = frac{2}{2} =1 \ x_{2}= frac{-(-7)+ sqrt{25} }{2*1} = frac{7+5}{2} = frac{12}{2} =6$