What are the solutions to the equation 11k^2+1k+1=4 ?

The solutions to the equation 11k^2+1k+1=4 are k=(-1+sqrt(133))/(22),k=(-1-sqrt(133))/(22)

Find the zeros of 11k^2+1k+1=4

The zeros of 11k^2+1k+1=4 are k=(-1+sqrt(133))/(22),k=(-1-sqrt(133))/(22)

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11 k^{2} + 1 k + 1 = 4

k = \frac{- 1 + 133}{22}, k = \frac{- 1 - 133}{22}

Decimal Notation

k = 0.47875 \dots, k = - 0.56966 \dots

Solution steps

11 k^{2} + 1 \cdot k + 1 = 4

Refine

11 k^{2} + k + 1 = 4

Move

4

to the left side

11 k^{2} + k - 3 = 0

Solve with the quadratic formula

k_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 11 ( - 3 )}{2 \cdot 11}

1^{2} - 4 \cdot 11 (- 3) = 133

k_{1, 2} = \frac{- 1 \pm 133}{2 \cdot 11}

Separate the solutions

k_{1} = \frac{- 1 + 133}{2 \cdot 11}, k_{2} = \frac{- 1 - 133}{2 \cdot 11}

k = \frac{- 1 + 133}{2 \cdot 11} : \frac{- 1 + 133}{22}

k = \frac{- 1 - 133}{2 \cdot 11} : \frac{- 1 - 133}{22}

The solutions to the quadratic equation are:

k = \frac{- 1 + 133}{22}, k = \frac{- 1 - 133}{22}

\frac{5 k}{6} \cdot \frac{3}{2 k ^{3}}

- 3 (x - 14) = 42

2 x \cdot 5 = \frac{x}{5} \cdot 5 - 13 \cdot 5

3 g - 5 = 2 g - 8

2 4^{2} + 4 5^{2} = x^{2}

What are the solutions to the equation 11k^2+1k+1=4 ?
The solutions to the equation 11k^2+1k+1=4 are k=(-1+sqrt(133))/(22),k=(-1-sqrt(133))/(22)
Find the zeros of 11k^2+1k+1=4
The zeros of 11k^2+1k+1=4 are k=(-1+sqrt(133))/(22),k=(-1-sqrt(133))/(22)