What are the solutions to the equation 9x^2+5x=1 ?

The solutions to the equation 9x^2+5x=1 are x=(-5+sqrt(61))/(18),x=(-5-sqrt(61))/(18)

Find the zeros of 9x^2+5x=1

The zeros of 9x^2+5x=1 are x=(-5+sqrt(61))/(18),x=(-5-sqrt(61))/(18)

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9 x^{2} + 5 x = 1

x = \frac{- 5 + 61}{18}, x = \frac{- 5 - 61}{18}

Decimal Notation

x = 0.15612 \dots, x = - 0.71168 \dots

Solution steps

9 x^{2} + 5 x = 1

Move

1

to the left side

9 x^{2} + 5 x - 1 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 9 ( - 1 )}{2 \cdot 9}

5^{2} - 4 \cdot 9 (- 1) = 61

x_{1, 2} = \frac{- 5 \pm 61}{2 \cdot 9}

Separate the solutions

x_{1} = \frac{- 5 + 61}{2 \cdot 9}, x_{2} = \frac{- 5 - 61}{2 \cdot 9}

x = \frac{- 5 + 61}{2 \cdot 9} : \frac{- 5 + 61}{18}

x = \frac{- 5 - 61}{2 \cdot 9} : \frac{- 5 - 61}{18}

The solutions to the quadratic equation are:

x = \frac{- 5 + 61}{18}, x = \frac{- 5 - 61}{18}

x^{2} - 6 x + 9

[1 + (\frac{1}{100})]^{100}

2 x^{2} + 5 x - 4 > - 2 x

2 x^{2} + 7 x < - 5

\frac{2 ^{x}}{2 ^{x + 1}}

What are the solutions to the equation 9x^2+5x=1 ?
The solutions to the equation 9x^2+5x=1 are x=(-5+sqrt(61))/(18),x=(-5-sqrt(61))/(18)
Find the zeros of 9x^2+5x=1
The zeros of 9x^2+5x=1 are x=(-5+sqrt(61))/(18),x=(-5-sqrt(61))/(18)