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

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

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

The zeros of quadraticformula 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}

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}

\frac{15}{42} \cdot (s im pl i f y)

x, - 8.5 = \frac{x}{2.5 - 3}

\frac{1}{3} \cdot (1800)

y = m x + b, x - y = - 15

1.5 \cdot 4 \frac{1}{5}

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