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

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

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

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

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

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

Decimal Notation

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

Solution steps

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

Write in the standard form

a x^{2} + b x + c = 0

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

Solve with the quadratic formula

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

1 2^{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 ( 2 - 1 )}{3}

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

The solutions to the quadratic equation are:

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

x + 5 = - 3 x + 25

(\frac{1}{x ^{2}}) - x + 2

- 3 (\frac{2}{3})^{0}

lo g_{\frac{1}{2}} (32)

3^{2}

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