What are the solutions to the equation 3x^2-33x-6=0 ?

The solutions to the equation 3x^2-33x-6=0 are x=(11+sqrt(129))/2 ,x=(11-sqrt(129))/2

Find the zeros of 3x^2-33x-6=0

The zeros of 3x^2-33x-6=0 are x=(11+sqrt(129))/2 ,x=(11-sqrt(129))/2

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3 x^{2} - 33 x - 6 = 0

x = \frac{11 + 129}{2}, x = \frac{11 - 129}{2}

Solution steps

3 x^{2} - 33 x - 6 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 33 ) \pm ( - 33 ) ^{2} - 4 \cdot 3 ( - 6 )}{2 \cdot 3}

(- 33)^{2} - 4 \cdot 3 (- 6) = 3129

x_{1, 2} = \frac{- ( - 33 ) \pm 3 129}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- ( - 33 ) + 3 129}{2 \cdot 3}, x_{2} = \frac{- ( - 33 ) - 3 129}{2 \cdot 3}

x = \frac{- ( - 33 ) + 3 129}{2 \cdot 3} : \frac{11 + 129}{2}

x = \frac{- ( - 33 ) - 3 129}{2 \cdot 3} : \frac{11 - 129}{2}

The solutions to the quadratic equation are:

x = \frac{11 + 129}{2}, x = \frac{11 - 129}{2}

(1 - x)^{3} - 3 x (1 - x)^{2}

3 x^{2} + 4 < 4 x

\frac{( x - 3 )}{( x + 4 )} \geq \frac{( x + 2 )}{( x - 5 )}

6 r^{3} + 8 r^{2} - 12 r - 16

3 x + 3 > 2 x + 7

What are the solutions to the equation 3x^2-33x-6=0 ?
The solutions to the equation 3x^2-33x-6=0 are x=(11+sqrt(129))/2 ,x=(11-sqrt(129))/2
Find the zeros of 3x^2-33x-6=0
The zeros of 3x^2-33x-6=0 are x=(11+sqrt(129))/2 ,x=(11-sqrt(129))/2