What are the solutions to the equation 3x^2+4x-5=0 ?

The solutions to the equation 3x^2+4x-5=0 are x=(-2+sqrt(19))/3 ,x=-(2+sqrt(19))/3

Find the zeros of 3x^2+4x-5=0

The zeros of 3x^2+4x-5=0 are x=(-2+sqrt(19))/3 ,x=-(2+sqrt(19))/3

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3 x^{2} + 4 x - 5 = 0

x = \frac{- 2 + 19}{3}, x = - \frac{2 + 19}{3}

Solution steps

3 x^{2} + 4 x - 5 = 0

Solve with the quadratic formula

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

4^{2} - 4 \cdot 3 (- 5) = 219

x_{1, 2} = \frac{- 4 \pm 2 19}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- 4 + 2 19}{2 \cdot 3}, x_{2} = \frac{- 4 - 2 19}{2 \cdot 3}

x = \frac{- 4 + 2 19}{2 \cdot 3} : \frac{- 2 + 19}{3}

x = \frac{- 4 - 2 19}{2 \cdot 3} : - \frac{2 + 19}{3}

The solutions to the quadratic equation are:

x = \frac{- 2 + 19}{3}, x = - \frac{2 + 19}{3}

3 w^{2} - 11 w = - 6

- 10 + 2 x - 4 x = - 24

2 x^{2} + 3 x - 20 > 0

3 (n - 7) = - 30

8 y - 3 - (5 - 2 y) = 4.3

What are the solutions to the equation 3x^2+4x-5=0 ?
The solutions to the equation 3x^2+4x-5=0 are x=(-2+sqrt(19))/3 ,x=-(2+sqrt(19))/3
Find the zeros of 3x^2+4x-5=0
The zeros of 3x^2+4x-5=0 are x=(-2+sqrt(19))/3 ,x=-(2+sqrt(19))/3