What are the solutions to the equation 5k^2+6k-6=0 ?

The solutions to the equation 5k^2+6k-6=0 are k=(-3+sqrt(39))/5 ,k=-(3+sqrt(39))/5

Find the zeros of 5k^2+6k-6=0

The zeros of 5k^2+6k-6=0 are k=(-3+sqrt(39))/5 ,k=-(3+sqrt(39))/5

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

k = \frac{- 3 + 39}{5}, k = - \frac{3 + 39}{5}

Decimal Notation

k = 0.64899 \dots, k = - 1.84899 \dots

Solution steps

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

Solve with the quadratic formula

k_{1, 2} = \frac{- 6 \pm 6 ^{2} - 4 \cdot 5 ( - 6 )}{2 \cdot 5}

6^{2} - 4 \cdot 5 (- 6) = 239

k_{1, 2} = \frac{- 6 \pm 2 39}{2 \cdot 5}

Separate the solutions

k_{1} = \frac{- 6 + 2 39}{2 \cdot 5}, k_{2} = \frac{- 6 - 2 39}{2 \cdot 5}

k = \frac{- 6 + 2 39}{2 \cdot 5} : \frac{- 3 + 39}{5}

k = \frac{- 6 - 2 39}{2 \cdot 5} : - \frac{3 + 39}{5}

The solutions to the quadratic equation are:

k = \frac{- 3 + 39}{5}, k = - \frac{3 + 39}{5}

8 p - 2 \geq 10 p - 3

3 (x - 5) = 1 + 5 (x - 6)

\frac{( - 8 q - 47 )}{( g + 7 )}

3 x^{2} - 11 x - 20 = 0

\frac{x ^{2} - 4}{x ^{2} + 36} \leq 0

What are the solutions to the equation 5k^2+6k-6=0 ?
The solutions to the equation 5k^2+6k-6=0 are k=(-3+sqrt(39))/5 ,k=-(3+sqrt(39))/5
Find the zeros of 5k^2+6k-6=0
The zeros of 5k^2+6k-6=0 are k=(-3+sqrt(39))/5 ,k=-(3+sqrt(39))/5