What are the solutions to the equation 5k^2+19k+11=2k ?

The solutions to the equation 5k^2+19k+11=2k are k=(-17+sqrt(69))/(10),k=(-17-sqrt(69))/(10)

Find the zeros of 5k^2+19k+11=2k

The zeros of 5k^2+19k+11=2k are k=(-17+sqrt(69))/(10),k=(-17-sqrt(69))/(10)

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5 k^{2} + 19 k + 11 = 2 k

k = \frac{- 17 + 69}{10}, k = \frac{- 17 - 69}{10}

Decimal Notation

k = - 0.86933 \dots, k = - 2.53066 \dots

Solution steps

5 k^{2} + 19 k + 11 = 2 k

Move

2 k

to the left side

5 k^{2} + 17 k + 11 = 0

Solve with the quadratic formula

k_{1, 2} = \frac{- 17 \pm 1 7 ^{2} - 4 \cdot 5 \cdot 11}{2 \cdot 5}

1 7^{2} - 4 \cdot 5 \cdot 11 = 69

k_{1, 2} = \frac{- 17 \pm 69}{2 \cdot 5}

Separate the solutions

k_{1} = \frac{- 17 + 69}{2 \cdot 5}, k_{2} = \frac{- 17 - 69}{2 \cdot 5}

k = \frac{- 17 + 69}{2 \cdot 5} : \frac{- 17 + 69}{10}

k = \frac{- 17 - 69}{2 \cdot 5} : \frac{- 17 - 69}{10}

The solutions to the quadratic equation are:

k = \frac{- 17 + 69}{10}, k = \frac{- 17 - 69}{10}

3 (3 x - 4)^{2} - 15 = 30

n = 5 (a - 3 h^{2})

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

r^{2} + r + 2 = 0

- 6 n^{2} - 8 n = 0

What are the solutions to the equation 5k^2+19k+11=2k ?
The solutions to the equation 5k^2+19k+11=2k are k=(-17+sqrt(69))/(10),k=(-17-sqrt(69))/(10)
Find the zeros of 5k^2+19k+11=2k
The zeros of 5k^2+19k+11=2k are k=(-17+sqrt(69))/(10),k=(-17-sqrt(69))/(10)