What are the solutions to the equation 2x^2+4x+1=5900 ?

The solutions to the equation 2x^2+4x+1=5900 are x=(-2+sqrt(11802))/2 ,x=-(2+sqrt(11802))/2

Find the zeros of 2x^2+4x+1=5900

The zeros of 2x^2+4x+1=5900 are x=(-2+sqrt(11802))/2 ,x=-(2+sqrt(11802))/2

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2 x^{2} + 4 x + 1 = 5900

x = \frac{- 2 + 11802}{2}, x = - \frac{2 + 11802}{2}

Solution steps

2 x^{2} + 4 x + 1 = 5900

Move

5900

to the left side

2 x^{2} + 4 x - 5899 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 2 ( - 5899 )}{2 \cdot 2}

4^{2} - 4 \cdot 2 (- 5899) = 211802

x_{1, 2} = \frac{- 4 \pm 2 11802}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 4 + 2 11802}{2 \cdot 2}, x_{2} = \frac{- 4 - 2 11802}{2 \cdot 2}

x = \frac{- 4 + 2 11802}{2 \cdot 2} : \frac{- 2 + 11802}{2}

x = \frac{- 4 - 2 11802}{2 \cdot 2} : - \frac{2 + 11802}{2}

The solutions to the quadratic equation are:

x = \frac{- 2 + 11802}{2}, x = - \frac{2 + 11802}{2}

2 k^{2} + 11 k + 12

\frac{4}{\frac{4}{5}} - \frac{4}{\frac{5}{4}}

4 x^{2} = 2 x + 6

∣ x + 2 ∣ + 6 \leq 11

(\frac{5}{6}) (\frac{2}{3})

What are the solutions to the equation 2x^2+4x+1=5900 ?
The solutions to the equation 2x^2+4x+1=5900 are x=(-2+sqrt(11802))/2 ,x=-(2+sqrt(11802))/2
Find the zeros of 2x^2+4x+1=5900
The zeros of 2x^2+4x+1=5900 are x=(-2+sqrt(11802))/2 ,x=-(2+sqrt(11802))/2