What are the solutions to the equation 9x^2-12x-4=0 ?

The solutions to the equation 9x^2-12x-4=0 are x=(2(1+sqrt(2)))/3 ,x=(2(1-sqrt(2)))/3

Find the zeros of 9x^2-12x-4=0

The zeros of 9x^2-12x-4=0 are x=(2(1+sqrt(2)))/3 ,x=(2(1-sqrt(2)))/3

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9 x^{2} - 12 x - 4 = 0

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

Decimal Notation

x = 1.60947 \dots, x = - 0.27614 \dots

Solution steps

9 x^{2} - 12 x - 4 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 12 ) \pm ( - 12 ) ^{2} - 4 \cdot 9 ( - 4 )}{2 \cdot 9}

(- 12)^{2} - 4 \cdot 9 (- 4) = 122

x_{1, 2} = \frac{- ( - 12 ) \pm 12 2}{2 \cdot 9}

Separate the solutions

x_{1} = \frac{- ( - 12 ) + 12 2}{2 \cdot 9}, x_{2} = \frac{- ( - 12 ) - 12 2}{2 \cdot 9}

x = \frac{- ( - 12 ) + 12 2}{2 \cdot 9} : \frac{2 ( 1 + 2 )}{3}

x = \frac{- ( - 12 ) - 12 2}{2 \cdot 9} : \frac{2 ( 1 - 2 )}{3}

The solutions to the quadratic equation are:

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

x^{2} - 9 x = - 2 x - 6

- \frac{1}{3} + \frac{1}{9}

(5^{- 6})^{- 4}

6 + 6 x - 10 = - 20

- 7 x + 6 x \leq - 4

What are the solutions to the equation 9x^2-12x-4=0 ?
The solutions to the equation 9x^2-12x-4=0 are x=(2(1+sqrt(2)))/3 ,x=(2(1-sqrt(2)))/3
Find the zeros of 9x^2-12x-4=0
The zeros of 9x^2-12x-4=0 are x=(2(1+sqrt(2)))/3 ,x=(2(1-sqrt(2)))/3