What are the solutions to the equation 3y^2+y=-1/12 ?

The solutions to the equation 3y^2+y=-1/12 are y=-1/6

Find the zeros of 3y^2+y=-1/12

The zeros of 3y^2+y=-1/12 are y=-1/6

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3 y^{2} + y = - \frac{1}{12}

y = - \frac{1}{6}

Decimal Notation

y = - 0.16666 \dots

Solution steps

3 y^{2} + y = - \frac{1}{12}

Multiply both sides by

12

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

Move

1

to the left side

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

Solve with the quadratic formula

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

1 2^{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}

x^{2} - 3 x - 20 = 0

x^{2} - 3 x - 21 = 0

\frac{1}{3} x^{2} = 44

6^{2} + 1 0^{2} = x^{2}

5 x^{2} + 3 x = 1

What are the solutions to the equation 3y^2+y=-1/12 ?
The solutions to the equation 3y^2+y=-1/12 are y=-1/6
Find the zeros of 3y^2+y=-1/12
The zeros of 3y^2+y=-1/12 are y=-1/6