What are the solutions to the equation 3x^2-4x-8=0 ?

The solutions to the equation 3x^2-4x-8=0 are x=(2(1+sqrt(7)))/3 ,x=(2(1-sqrt(7)))/3

Find the zeros of 3x^2-4x-8=0

The zeros of 3x^2-4x-8=0 are x=(2(1+sqrt(7)))/3 ,x=(2(1-sqrt(7)))/3

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3 x^{2} - 4 x - 8 = 0

x = \frac{2 ( 1 + 7 )}{3}, x = \frac{2 ( 1 - 7 )}{3}

Decimal Notation

x = 2.43050 \dots, x = - 1.09716 \dots

Solution steps

3 x^{2} - 4 x - 8 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 3 ( - 8 )}{2 \cdot 3}

(- 4)^{2} - 4 \cdot 3 (- 8) = 47

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

Separate the solutions

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

x = \frac{- ( - 4 ) + 4 7}{2 \cdot 3} : \frac{2 ( 1 + 7 )}{3}

x = \frac{- ( - 4 ) - 4 7}{2 \cdot 3} : \frac{2 ( 1 - 7 )}{3}

The solutions to the quadratic equation are:

x = \frac{2 ( 1 + 7 )}{3}, x = \frac{2 ( 1 - 7 )}{3}

8 (2 y + 5) = 9 y + 12

ln (\frac{x y}{z})

x^{2} - x - 3 = 0

9 w^{2} + 9 w + 2

\frac{( x ^{3} + y ^{3} )}{x - y}

What are the solutions to the equation 3x^2-4x-8=0 ?
The solutions to the equation 3x^2-4x-8=0 are x=(2(1+sqrt(7)))/3 ,x=(2(1-sqrt(7)))/3
Find the zeros of 3x^2-4x-8=0
The zeros of 3x^2-4x-8=0 are x=(2(1+sqrt(7)))/3 ,x=(2(1-sqrt(7)))/3