What are the solutions to the equation 9k^2+6k=2 ?

The solutions to the equation 9k^2+6k=2 are k=(-1+sqrt(3))/3 ,k=-(1+sqrt(3))/3

Find the zeros of 9k^2+6k=2

The zeros of 9k^2+6k=2 are k=(-1+sqrt(3))/3 ,k=-(1+sqrt(3))/3

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9 k^{2} + 6 k = 2

k = \frac{- 1 + 3}{3}, k = - \frac{1 + 3}{3}

Decimal Notation

k = 0.24401 \dots, k = - 0.91068 \dots

Solution steps

9 k^{2} + 6 k = 2

Move

2

to the left side

9 k^{2} + 6 k - 2 = 0

Solve with the quadratic formula

k_{1, 2} = \frac{- 6 \pm 6 ^{2} - 4 \cdot 9 ( - 2 )}{2 \cdot 9}

6^{2} - 4 \cdot 9 (- 2) = 63

k_{1, 2} = \frac{- 6 \pm 6 3}{2 \cdot 9}

Separate the solutions

k_{1} = \frac{- 6 + 6 3}{2 \cdot 9}, k_{2} = \frac{- 6 - 6 3}{2 \cdot 9}

k = \frac{- 6 + 6 3}{2 \cdot 9} : \frac{- 1 + 3}{3}

k = \frac{- 6 - 6 3}{2 \cdot 9} : - \frac{1 + 3}{3}

The solutions to the quadratic equation are:

k = \frac{- 1 + 3}{3}, k = - \frac{1 + 3}{3}

3 x^{2} y + 7 x y^{2} + 2 x^{2} y - 9 x y^{2}

(5 + 13)^{2} = 230

\frac{3}{4} (8 x - 6) - 2 = \frac{1}{2} (1 - 2 x)

- 6 a^{2} - 25 a - 25

5 x^{2} - 30 x = 0

What are the solutions to the equation 9k^2+6k=2 ?
The solutions to the equation 9k^2+6k=2 are k=(-1+sqrt(3))/3 ,k=-(1+sqrt(3))/3
Find the zeros of 9k^2+6k=2
The zeros of 9k^2+6k=2 are k=(-1+sqrt(3))/3 ,k=-(1+sqrt(3))/3