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

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

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

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

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

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

Decimal Notation

x = 1.13504 \dots, x = - 1.46837 \dots

Solution steps

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

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

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

x^{2} - 8 x - 13 = 0

∣ - 6 - 9 i ∣

- 3 x + 2 > 2 x + 3

x - 2 = 2 - x

- 7 + x = - 6

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