What are the solutions to the equation 9x^2+4x=1 ?

The solutions to the equation 9x^2+4x=1 are x=(-2+sqrt(13))/9 ,x=-(2+sqrt(13))/9

Find the zeros of 9x^2+4x=1

The zeros of 9x^2+4x=1 are x=(-2+sqrt(13))/9 ,x=-(2+sqrt(13))/9

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9 x^{2} + 4 x = 1

x = \frac{- 2 + 13}{9}, x = - \frac{2 + 13}{9}

Decimal Notation

x = 0.17839 \dots, x = - 0.62283 \dots

Solution steps

9 x^{2} + 4 x = 1

Move

1

to the left side

9 x^{2} + 4 x - 1 = 0

Solve with the quadratic formula

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

4^{2} - 4 \cdot 9 (- 1) = 213

x_{1, 2} = \frac{- 4 \pm 2 13}{2 \cdot 9}

Separate the solutions

x_{1} = \frac{- 4 + 2 13}{2 \cdot 9}, x_{2} = \frac{- 4 - 2 13}{2 \cdot 9}

x = \frac{- 4 + 2 13}{2 \cdot 9} : \frac{- 2 + 13}{9}

x = \frac{- 4 - 2 13}{2 \cdot 9} : - \frac{2 + 13}{9}

The solutions to the quadratic equation are:

x = \frac{- 2 + 13}{9}, x = - \frac{2 + 13}{9}

x (x + 5) = 104

2 x^{2} + b x + 3 = 0

x^{2} + 5.71 \cdot 1 0^{- 12} x - 5.71 \cdot 1 0^{- 10} = 0

a, x^{2} + y^{2} - x y = a^{2}

y^{2} - 2 y + 1 = y + 1

What are the solutions to the equation 9x^2+4x=1 ?
The solutions to the equation 9x^2+4x=1 are x=(-2+sqrt(13))/9 ,x=-(2+sqrt(13))/9
Find the zeros of 9x^2+4x=1
The zeros of 9x^2+4x=1 are x=(-2+sqrt(13))/9 ,x=-(2+sqrt(13))/9