What are the solutions to the equation 3y^2-12y+12=0 ?

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

Find the zeros of 3y^2-12y+12=0

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

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3 y^{2} - 12 y + 12 = 0

y = 2

Solution steps

3 y^{2} - 12 y + 12 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- ( - 12 ) \pm ( - 12 ) ^{2} - 4 \cdot 3 \cdot 12}{2 \cdot 3}

(- 12)^{2} - 4 \cdot 3 \cdot 12 = 0

y_{1, 2} = \frac{- ( - 12 ) \pm 0}{2 \cdot 3}

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

\frac{- ( - 12 )}{2 \cdot 3} = 2

y = 2

The solution to the quadratic equation is:

y = 2

2 2^{2}

28 n^{2} - 184 n - 96

(\frac{4 2 ^{\frac{1}{3}}}{6 ^{\frac{1}{3}}})^{2}

m + 9 < 8

- 5 x - 15 = x^{2} + x - 2

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