What are the solutions to the equation 9x^2+4x-8=0 ?

The solutions to the equation 9x^2+4x-8=0 are x=(2(sqrt(19)-1))/9 ,x=-(2(1+sqrt(19)))/9

Find the zeros of 9x^2+4x-8=0

The zeros of 9x^2+4x-8=0 are x=(2(sqrt(19)-1))/9 ,x=-(2(1+sqrt(19)))/9

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

x = \frac{2 ( 19 - 1 )}{9}, x = - \frac{2 ( 1 + 19 )}{9}

Decimal Notation

x = 0.74642 \dots, x = - 1.19086 \dots

Solution steps

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

Solve with the quadratic formula

x_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 9 ( - 8 )}{2 \cdot 9}

4^{2} - 4 \cdot 9 (- 8) = 419

x_{1, 2} = \frac{- 4 \pm 4 19}{2 \cdot 9}

Separate the solutions

x_{1} = \frac{- 4 + 4 19}{2 \cdot 9}, x_{2} = \frac{- 4 - 4 19}{2 \cdot 9}

x = \frac{- 4 + 4 19}{2 \cdot 9} : \frac{2 ( 19 - 1 )}{9}

x = \frac{- 4 - 4 19}{2 \cdot 9} : - \frac{2 ( 1 + 19 )}{9}

The solutions to the quadratic equation are:

x = \frac{2 ( 19 - 1 )}{9}, x = - \frac{2 ( 1 + 19 )}{9}

15 + x (x + 2 x) = 10

2 r^{2} - r + 1 = 0

b^{2} + b - 20 = 0

c^{2} - 6 c - 7 = 0

x, 3 x^{2} - 4 x y + 2 y^{2} = 17

What are the solutions to the equation 9x^2+4x-8=0 ?
The solutions to the equation 9x^2+4x-8=0 are x=(2(sqrt(19)-1))/9 ,x=-(2(1+sqrt(19)))/9
Find the zeros of 9x^2+4x-8=0
The zeros of 9x^2+4x-8=0 are x=(2(sqrt(19)-1))/9 ,x=-(2(1+sqrt(19)))/9