What are the solutions to the equation 2x^2+9x+1=0 ?

The solutions to the equation 2x^2+9x+1=0 are x=(-9+sqrt(73))/4 ,x=(-9-sqrt(73))/4

Find the zeros of 2x^2+9x+1=0

The zeros of 2x^2+9x+1=0 are x=(-9+sqrt(73))/4 ,x=(-9-sqrt(73))/4

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

x = \frac{- 9 + 73}{4}, x = \frac{- 9 - 73}{4}

Decimal Notation

x = - 0.11399 \dots, x = - 4.38600 \dots

Solution steps

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

Solve with the quadratic formula

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

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

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

Separate the solutions

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

x = \frac{- 9 + 73}{2 \cdot 2} : \frac{- 9 + 73}{4}

x = \frac{- 9 - 73}{2 \cdot 2} : \frac{- 9 - 73}{4}

The solutions to the quadratic equation are:

x = \frac{- 9 + 73}{4}, x = \frac{- 9 - 73}{4}

t^{2} = - 2 t

- 2 (v + 2) + 7 = 3 (v + 6)

10 + 6 x = 15 + 9 x - 3 x

x^{2} - 11 x = 0

\frac{6 x}{\frac{4}{y} - \frac{3}{x}}

What are the solutions to the equation 2x^2+9x+1=0 ?
The solutions to the equation 2x^2+9x+1=0 are x=(-9+sqrt(73))/4 ,x=(-9-sqrt(73))/4
Find the zeros of 2x^2+9x+1=0
The zeros of 2x^2+9x+1=0 are x=(-9+sqrt(73))/4 ,x=(-9-sqrt(73))/4