What are the solutions to the equation 2x^2+5x+1=0 ?

The solutions to the equation 2x^2+5x+1=0 are x=(-5+sqrt(17))/4 ,x=(-5-sqrt(17))/4

Find the zeros of 2x^2+5x+1=0

The zeros of 2x^2+5x+1=0 are x=(-5+sqrt(17))/4 ,x=(-5-sqrt(17))/4

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

x = \frac{- 5 + 17}{4}, x = \frac{- 5 - 17}{4}

Decimal Notation

x = - 0.21922 \dots, x = - 2.28077 \dots

Solution steps

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

Solve with the quadratic formula

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

5^{2} - 4 \cdot 2 \cdot 1 = 17

x_{1, 2} = \frac{- 5 \pm 17}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 5 + 17}{2 \cdot 2}, x_{2} = \frac{- 5 - 17}{2 \cdot 2}

x = \frac{- 5 + 17}{2 \cdot 2} : \frac{- 5 + 17}{4}

x = \frac{- 5 - 17}{2 \cdot 2} : \frac{- 5 - 17}{4}

The solutions to the quadratic equation are:

x = \frac{- 5 + 17}{4}, x = \frac{- 5 - 17}{4}

0.001 \cdot R < 12

∣ 4 - x ∣ \leq 0

(\frac{2}{3}) 4

- (6 h + 7) + 8 = - 19

1000 e^{0.01 (4)}

What are the solutions to the equation 2x^2+5x+1=0 ?
The solutions to the equation 2x^2+5x+1=0 are x=(-5+sqrt(17))/4 ,x=(-5-sqrt(17))/4
Find the zeros of 2x^2+5x+1=0
The zeros of 2x^2+5x+1=0 are x=(-5+sqrt(17))/4 ,x=(-5-sqrt(17))/4