What are the solutions to the equation (12y-2((18-3y)/4)y-3y^2)=0 ?

The solutions to the equation (12y-2((18-3y)/4)y-3y^2)=0 are y=0,y=2

Find the zeros of (12y-2((18-3y)/4)y-3y^2)=0

The zeros of (12y-2((18-3y)/4)y-3y^2)=0 are y=0,y=2

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(12 y - 2 (\frac{18 - 3 y}{4}) y - 3 y^{2}) = 0

y = 0, y = 2

Solution steps

(12 y - 2 (\frac{18 - 3 y}{4}) y - 3 y^{2}) = 0

Factor

12 y - 2 (\frac{18 - 3 y}{4}) y - 3 y^{2} : y (- \frac{3 ( y + 6 )}{2} + 12)

y (- \frac{3 ( y + 6 )}{2} + 12) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

y = 0 or - \frac{3 ( y + 6 )}{2} + 12 = 0

Solve

- \frac{3 ( y + 6 )}{2} + 12 = 0 : y = 2

The solutions to the quadratic equation are:

y = 0, y = 2

(3 x - 1) (x - 2) = 22

2 x^{2} - 9 x = 0

x, x^{2} y + 2 y^{3} = 3 x + 2 y

4 n^{2} - 8 n + 90 = 9

x, h + g = x^{2}

What are the solutions to the equation (12y-2((18-3y)/4)y-3y^2)=0 ?
The solutions to the equation (12y-2((18-3y)/4)y-3y^2)=0 are y=0,y=2
Find the zeros of (12y-2((18-3y)/4)y-3y^2)=0
The zeros of (12y-2((18-3y)/4)y-3y^2)=0 are y=0,y=2