What are the solutions to the equation 4y^2=12y ?

The solutions to the equation 4y^2=12y are y=0,y=3

Find the zeros of 4y^2=12y

The zeros of 4y^2=12y are y=0,y=3

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4 y^{2} = 12 y

y = 0, y = 3

Solution steps

4 y^{2} = 12 y

Move

12 y

to the left side

4 y^{2} - 12 y = 0

Factor

4 y^{2} - 12 y : 4 y (y - 3)

4 y (y - 3) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

y = 0 or y - 3 = 0

Solve

y - 3 = 0 : y = 3

The solutions to the quadratic equation are:

y = 0, y = 3

(5 x - 3) (x - 5) = 0

x^{2} - 8 x + 3 = 2

2 x - 2 = x^{2} + 7 x - 38

x^{2} - 10 x = - 41

x, 9 x^{2} + 3 x - 1 = 0

What are the solutions to the equation 4y^2=12y ?
The solutions to the equation 4y^2=12y are y=0,y=3
Find the zeros of 4y^2=12y
The zeros of 4y^2=12y are y=0,y=3