What are the solutions to the equation -3x^2-18x+2=0 ?

The solutions to the equation -3x^2-18x+2=0 are x=-(9+sqrt(87))/3 ,x=(sqrt(87)-9)/3

Find the zeros of -3x^2-18x+2=0

The zeros of -3x^2-18x+2=0 are x=-(9+sqrt(87))/3 ,x=(sqrt(87)-9)/3

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- 3 x^{2} - 18 x + 2 = 0

x = - \frac{9 + 87}{3}, x = \frac{87 - 9}{3}

Decimal Notation

x = - 6.10912 \dots, x = 0.10912 \dots

Solution steps

- 3 x^{2} - 18 x + 2 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 18 ) \pm ( - 18 ) ^{2} - 4 ( - 3 ) \cdot 2}{2 ( - 3 )}

(- 18)^{2} - 4 (- 3) \cdot 2 = 287

x_{1, 2} = \frac{- ( - 18 ) \pm 2 87}{2 ( - 3 )}

Separate the solutions

x_{1} = \frac{- ( - 18 ) + 2 87}{2 ( - 3 )}, x_{2} = \frac{- ( - 18 ) - 2 87}{2 ( - 3 )}

x = \frac{- ( - 18 ) + 2 87}{2 ( - 3 )} : - \frac{9 + 87}{3}

x = \frac{- ( - 18 ) - 2 87}{2 ( - 3 )} : \frac{87 - 9}{3}

The solutions to the quadratic equation are:

x = - \frac{9 + 87}{3}, x = \frac{87 - 9}{3}

\frac{1}{1 + \frac{1}{1 + \frac{1}{1 + \frac{1}{3}}}}

x^{2} \leq x + 20

x^{\frac{2}{3}} \cdot x^{\frac{4}{3}}

- 3 + \frac{x}{4} > 5

\frac{1}{x} \leq 2

What are the solutions to the equation -3x^2-18x+2=0 ?
The solutions to the equation -3x^2-18x+2=0 are x=-(9+sqrt(87))/3 ,x=(sqrt(87)-9)/3
Find the zeros of -3x^2-18x+2=0
The zeros of -3x^2-18x+2=0 are x=-(9+sqrt(87))/3 ,x=(sqrt(87)-9)/3