What are the solutions to the equation 3x^2+2x-26=0 ?

The solutions to the equation 3x^2+2x-26=0 are x=(-1+sqrt(79))/3 ,x=-(1+sqrt(79))/3

Find the zeros of 3x^2+2x-26=0

The zeros of 3x^2+2x-26=0 are x=(-1+sqrt(79))/3 ,x=-(1+sqrt(79))/3

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3 x^{2} + 2 x - 26 = 0

x = \frac{- 1 + 79}{3}, x = - \frac{1 + 79}{3}

Decimal Notation

x = 2.62939 \dots, x = - 3.29606 \dots

Solution steps

3 x^{2} + 2 x - 26 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 3 ( - 26 )}{2 \cdot 3}

2^{2} - 4 \cdot 3 (- 26) = 279

x_{1, 2} = \frac{- 2 \pm 2 79}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- 2 + 2 79}{2 \cdot 3}, x_{2} = \frac{- 2 - 2 79}{2 \cdot 3}

x = \frac{- 2 + 2 79}{2 \cdot 3} : \frac{- 1 + 79}{3}

x = \frac{- 2 - 2 79}{2 \cdot 3} : - \frac{1 + 79}{3}

The solutions to the quadratic equation are:

x = \frac{- 1 + 79}{3}, x = - \frac{1 + 79}{3}

- 41 a - 4 = 47 a - 3

\frac{3 ^{6}}{3 ^{2}}

- 2 ∣ x - 1 ∣ + 8 = - 6

(\frac{3 x}{2})^{2}

\frac{2 x}{3} - 4 (\frac{x}{5} - \frac{1}{6}) = \frac{1}{15}

What are the solutions to the equation 3x^2+2x-26=0 ?
The solutions to the equation 3x^2+2x-26=0 are x=(-1+sqrt(79))/3 ,x=-(1+sqrt(79))/3
Find the zeros of 3x^2+2x-26=0
The zeros of 3x^2+2x-26=0 are x=(-1+sqrt(79))/3 ,x=-(1+sqrt(79))/3