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(sqrt(7)-1))/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(sqrt(7)-1))/3 ,x=-(2(1+sqrt(7)))/3

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

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

Decimal

x = 1.09716 \dots, x = - 2.43050 \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 ( 7 - 1 )}{3}

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

The solutions to the quadratic equation are:

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

- 4 k \geq k + 15

36 x - x^{2} = 62

\frac{( 2 x ^{2} y ) ^{- 1}}{( - 2 x ^{2} y ^{3} ) ^{2}}

125 - w^{3}

12 x + 7 < - 11 or 5 x - 8 > 40

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(sqrt(7)-1))/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(sqrt(7)-1))/3 ,x=-(2(1+sqrt(7)))/3