What are the solutions to the equation 36y^2-12y+1=0 ?

The solutions to the equation 36y^2-12y+1=0 are y= 1/6

Find the zeros of 36y^2-12y+1=0

The zeros of 36y^2-12y+1=0 are y= 1/6

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36 y^{2} - 12 y + 1 = 0

y = \frac{1}{6}

Decimal Notation

y = 0.16666 \dots

Solution steps

36 y^{2} - 12 y + 1 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- ( - 12 ) \pm ( - 12 ) ^{2} - 4 \cdot 36 \cdot 1}{2 \cdot 36}

(- 12)^{2} - 4 \cdot 36 \cdot 1 = 0

y_{1, 2} = \frac{- ( - 12 ) \pm 0}{2 \cdot 36}

y = \frac{- ( - 12 )}{2 \cdot 36}

\frac{- ( - 12 )}{2 \cdot 36} = \frac{1}{6}

y = \frac{1}{6}

The solution to the quadratic equation is:

y = \frac{1}{6}

z^{3} + 9 z^{2} + 20 z

lo g_{10} (∣ x - 1 ∣) = 3

2^{2 x} - 5 (2^{x}) + 4 < 0

5 x - 2 = 4 x

9 x^{2} - 60 x + 100

What are the solutions to the equation 36y^2-12y+1=0 ?
The solutions to the equation 36y^2-12y+1=0 are y= 1/6
Find the zeros of 36y^2-12y+1=0
The zeros of 36y^2-12y+1=0 are y= 1/6