What are the solutions to the equation 3x^2-6x-7=0 ?

The solutions to the equation 3x^2-6x-7=0 are x=(3+sqrt(30))/3 ,x=(3-sqrt(30))/3

Find the zeros of 3x^2-6x-7=0

The zeros of 3x^2-6x-7=0 are x=(3+sqrt(30))/3 ,x=(3-sqrt(30))/3

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3 x^{2} - 6 x - 7 = 0

x = \frac{3 + 30}{3}, x = \frac{3 - 30}{3}

Decimal Notation

x = 2.82574 \dots, x = - 0.82574 \dots

Solution steps

3 x^{2} - 6 x - 7 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 6 ) \pm ( - 6 ) ^{2} - 4 \cdot 3 ( - 7 )}{2 \cdot 3}

(- 6)^{2} - 4 \cdot 3 (- 7) = 230

x_{1, 2} = \frac{- ( - 6 ) \pm 2 30}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- ( - 6 ) + 2 30}{2 \cdot 3}, x_{2} = \frac{- ( - 6 ) - 2 30}{2 \cdot 3}

x = \frac{- ( - 6 ) + 2 30}{2 \cdot 3} : \frac{3 + 30}{3}

x = \frac{- ( - 6 ) - 2 30}{2 \cdot 3} : \frac{3 - 30}{3}

The solutions to the quadratic equation are:

x = \frac{3 + 30}{3}, x = \frac{3 - 30}{3}

z^{10} + 17 \geq 16

18 - 6.3 x = 49.5

\frac{3}{5} \cdot \frac{1}{7}

x (1 - x^{2})^{3} > 5 (1 - x^{2})^{3}

\frac{3 + 3}{4 12}

What are the solutions to the equation 3x^2-6x-7=0 ?
The solutions to the equation 3x^2-6x-7=0 are x=(3+sqrt(30))/3 ,x=(3-sqrt(30))/3
Find the zeros of 3x^2-6x-7=0
The zeros of 3x^2-6x-7=0 are x=(3+sqrt(30))/3 ,x=(3-sqrt(30))/3