What are the solutions to the equation 4y^2-12y=0 ?

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

Find the zeros of 4y^2-12y=0

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

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4 y^{2} - 12 y = 0

y = 3, y = 0

Solution steps

4 y^{2} - 12 y = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- ( - 12 ) \pm ( - 12 ) ^{2} - 4 \cdot 4 \cdot 0}{2 \cdot 4}

(- 12)^{2} - 4 \cdot 4 \cdot 0 = 12

y_{1, 2} = \frac{- ( - 12 ) \pm 12}{2 \cdot 4}

Separate the solutions

y_{1} = \frac{- ( - 12 ) + 12}{2 \cdot 4}, y_{2} = \frac{- ( - 12 ) - 12}{2 \cdot 4}

y = \frac{- ( - 12 ) + 12}{2 \cdot 4} : 3

y = \frac{- ( - 12 ) - 12}{2 \cdot 4} : 0

The solutions to the quadratic equation are:

y = 3, y = 0

r^{2} + 3 r = 8

x^{2} + 8 x + 16 = 0

2 x^{2} - 50 x = 0

3 r^{2} - 8 = 6

- 3 x^{2} - 2 x + 16 = 0

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