What are the solutions to the equation 3x^2+11x-7=0 ?

The solutions to the equation 3x^2+11x-7=0 are x=(-11+sqrt(205))/6 ,x=(-11-sqrt(205))/6

Find the zeros of 3x^2+11x-7=0

The zeros of 3x^2+11x-7=0 are x=(-11+sqrt(205))/6 ,x=(-11-sqrt(205))/6

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3 x^{2} + 11 x - 7 = 0

x = \frac{- 11 + 205}{6}, x = \frac{- 11 - 205}{6}

Decimal Notation

x = 0.55297 \dots, x = - 4.21963 \dots

Solution steps

3 x^{2} + 11 x - 7 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 11 \pm 1 1 ^{2} - 4 \cdot 3 ( - 7 )}{2 \cdot 3}

1 1^{2} - 4 \cdot 3 (- 7) = 205

x_{1, 2} = \frac{- 11 \pm 205}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- 11 + 205}{2 \cdot 3}, x_{2} = \frac{- 11 - 205}{2 \cdot 3}

x = \frac{- 11 + 205}{2 \cdot 3} : \frac{- 11 + 205}{6}

x = \frac{- 11 - 205}{2 \cdot 3} : \frac{- 11 - 205}{6}

The solutions to the quadratic equation are:

x = \frac{- 11 + 205}{6}, x = \frac{- 11 - 205}{6}

A = x^{2} + x - 2

8 x^{2} - 28 x - 60 = 0

r^{2} + \frac{4}{3} r = \frac{4}{3}

c, y^{2} + 2 x^{2} = c^{2}

2 = (1 - 2 (1) + 5) t^{2} + (2 (1) - 2 (1)) t + 1

What are the solutions to the equation 3x^2+11x-7=0 ?
The solutions to the equation 3x^2+11x-7=0 are x=(-11+sqrt(205))/6 ,x=(-11-sqrt(205))/6
Find the zeros of 3x^2+11x-7=0
The zeros of 3x^2+11x-7=0 are x=(-11+sqrt(205))/6 ,x=(-11-sqrt(205))/6