What are the solutions to the equation 3k^2-4k-4=k ?

The solutions to the equation 3k^2-4k-4=k are k=(5+sqrt(73))/6 ,k=(5-sqrt(73))/6

Find the zeros of 3k^2-4k-4=k

The zeros of 3k^2-4k-4=k are k=(5+sqrt(73))/6 ,k=(5-sqrt(73))/6

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3 k^{2} - 4 k - 4 = k

k = \frac{5 + 73}{6}, k = \frac{5 - 73}{6}

Solution steps

3 k^{2} - 4 k - 4 = k

Move

k

to the left side

3 k^{2} - 5 k - 4 = 0

Solve with the quadratic formula

k_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 3 ( - 4 )}{2 \cdot 3}

(- 5)^{2} - 4 \cdot 3 (- 4) = 73

k_{1, 2} = \frac{- ( - 5 ) \pm 73}{2 \cdot 3}

Separate the solutions

k_{1} = \frac{- ( - 5 ) + 73}{2 \cdot 3}, k_{2} = \frac{- ( - 5 ) - 73}{2 \cdot 3}

k = \frac{- ( - 5 ) + 73}{2 \cdot 3} : \frac{5 + 73}{6}

k = \frac{- ( - 5 ) - 73}{2 \cdot 3} : \frac{5 - 73}{6}

The solutions to the quadratic equation are:

k = \frac{5 + 73}{6}, k = \frac{5 - 73}{6}

(3 x + 9)^{2} = 81

k (x) = 2 x^{2} + 3 x - 1, f (x, h) (a + h)

(3 x - 2)^{2} = (2 x + 2)^{2} + x^{2}

2 m^{2} - 8 m = 0

4 - 2 x + 3 x^{2} = 0

What are the solutions to the equation 3k^2-4k-4=k ?
The solutions to the equation 3k^2-4k-4=k are k=(5+sqrt(73))/6 ,k=(5-sqrt(73))/6
Find the zeros of 3k^2-4k-4=k
The zeros of 3k^2-4k-4=k are k=(5+sqrt(73))/6 ,k=(5-sqrt(73))/6