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

The solutions to the equation 9x^2+8x-8=0 are x=(2(sqrt(22)-2))/9 ,x=-(2(2+sqrt(22)))/9

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

The zeros of 9x^2+8x-8=0 are x=(2(sqrt(22)-2))/9 ,x=-(2(2+sqrt(22)))/9

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

x = \frac{2 ( 22 - 2 )}{9}, x = - \frac{2 ( 2 + 22 )}{9}

Decimal Notation

x = 0.59787 \dots, x = - 1.48675 \dots

Solution steps

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

Solve with the quadratic formula

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

8^{2} - 4 \cdot 9 (- 8) = 422

x_{1, 2} = \frac{- 8 \pm 4 22}{2 \cdot 9}

Separate the solutions

x_{1} = \frac{- 8 + 4 22}{2 \cdot 9}, x_{2} = \frac{- 8 - 4 22}{2 \cdot 9}

x = \frac{- 8 + 4 22}{2 \cdot 9} : \frac{2 ( 22 - 2 )}{9}

x = \frac{- 8 - 4 22}{2 \cdot 9} : - \frac{2 ( 2 + 22 )}{9}

The solutions to the quadratic equation are:

x = \frac{2 ( 22 - 2 )}{9}, x = - \frac{2 ( 2 + 22 )}{9}

(\frac{1}{e})^{- 3}

r^{2} + 7 r + 10

\frac{1}{8} + \frac{5}{8}

\frac{x}{4} - \frac{x}{3} = - 1 - \frac{x}{5}

\frac{2}{5 - 2}

What are the solutions to the equation 9x^2+8x-8=0 ?
The solutions to the equation 9x^2+8x-8=0 are x=(2(sqrt(22)-2))/9 ,x=-(2(2+sqrt(22)))/9
Find the zeros of 9x^2+8x-8=0
The zeros of 9x^2+8x-8=0 are x=(2(sqrt(22)-2))/9 ,x=-(2(2+sqrt(22)))/9