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

The solutions to the equation 9x^2+x=2 are x=(-1+sqrt(73))/(18),x=(-1-sqrt(73))/(18)

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

The zeros of 9x^2+x=2 are x=(-1+sqrt(73))/(18),x=(-1-sqrt(73))/(18)

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9 x^{2} + x = 2

x = \frac{- 1 + 73}{18}, x = \frac{- 1 - 73}{18}

Decimal Notation

x = 0.41911 \dots, x = - 0.53022 \dots

Solution steps

9 x^{2} + x = 2

Move

2

to the left side

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

Solve with the quadratic formula

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

1^{2} - 4 \cdot 9 (- 2) = 73

x_{1, 2} = \frac{- 1 \pm 73}{2 \cdot 9}

Separate the solutions

x_{1} = \frac{- 1 + 73}{2 \cdot 9}, x_{2} = \frac{- 1 - 73}{2 \cdot 9}

x = \frac{- 1 + 73}{2 \cdot 9} : \frac{- 1 + 73}{18}

x = \frac{- 1 - 73}{2 \cdot 9} : \frac{- 1 - 73}{18}

The solutions to the quadratic equation are:

x = \frac{- 1 + 73}{18}, x = \frac{- 1 - 73}{18}

\frac{x y}{x y}

∣ x ∣ \leq 1

x^{3} - 2 x^{2} < x^{3} - 2 x^{2} - 3 x + 3

44 = - 4 + 8 p

550 t - 4.9 t^{2} + 10 = 0

What are the solutions to the equation 9x^2+x=2 ?
The solutions to the equation 9x^2+x=2 are x=(-1+sqrt(73))/(18),x=(-1-sqrt(73))/(18)
Find the zeros of 9x^2+x=2
The zeros of 9x^2+x=2 are x=(-1+sqrt(73))/(18),x=(-1-sqrt(73))/(18)