What are the solutions to the equation 3y^2+y-5=0 ?

The solutions to the equation 3y^2+y-5=0 are y=(-1+sqrt(61))/6 ,y=(-1-sqrt(61))/6

Find the zeros of 3y^2+y-5=0

The zeros of 3y^2+y-5=0 are y=(-1+sqrt(61))/6 ,y=(-1-sqrt(61))/6

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3 y^{2} + y - 5 = 0

y = \frac{- 1 + 61}{6}, y = \frac{- 1 - 61}{6}

Solution steps

3 y^{2} + y - 5 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 3 ( - 5 )}{2 \cdot 3}

1^{2} - 4 \cdot 3 (- 5) = 61

y_{1, 2} = \frac{- 1 \pm 61}{2 \cdot 3}

Separate the solutions

y_{1} = \frac{- 1 + 61}{2 \cdot 3}, y_{2} = \frac{- 1 - 61}{2 \cdot 3}

y = \frac{- 1 + 61}{2 \cdot 3} : \frac{- 1 + 61}{6}

y = \frac{- 1 - 61}{2 \cdot 3} : \frac{- 1 - 61}{6}

The solutions to the quadratic equation are:

y = \frac{- 1 + 61}{6}, y = \frac{- 1 - 61}{6}

76 - (- 7 + 5 (9 - 14 + 7) - 5) - 4 (- 3)

14 p^{2} + 14 p

\frac{5 x ^{8} y ^{4}}{x ^{6} y ^{5}}

34 m = 11

5 x^{2} - 6 x = 0

What are the solutions to the equation 3y^2+y-5=0 ?
The solutions to the equation 3y^2+y-5=0 are y=(-1+sqrt(61))/6 ,y=(-1-sqrt(61))/6
Find the zeros of 3y^2+y-5=0
The zeros of 3y^2+y-5=0 are y=(-1+sqrt(61))/6 ,y=(-1-sqrt(61))/6