What are the solutions to the equation k^2-31-2k=-6-3k^2-2k ?

The solutions to the equation k^2-31-2k=-6-3k^2-2k are k= 5/2 ,k=-5/2

Find the zeros of k^2-31-2k=-6-3k^2-2k

The zeros of k^2-31-2k=-6-3k^2-2k are k= 5/2 ,k=-5/2

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k^{2} - 31 - 2 k = - 6 - 3 k^{2} - 2 k

k = \frac{5}{2}, k = - \frac{5}{2}

Decimal Notation

k = 2.5, k = - 2.5

Solution steps

k^{2} - 31 - 2 k = - 6 - 3 k^{2} - 2 k

Move

31

to the right side

k^{2} - 2 k = - 3 k^{2} - 2 k + 25

Move

2 k

to the left side

k^{2} = - 3 k^{2} + 25

Move

3 k^{2}

to the left side

4 k^{2} = 25

Divide both sides by

4

k^{2} = \frac{25}{4}

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

k = \frac{25}{4}, k = - \frac{25}{4}

\frac{25}{4} = \frac{5}{2}

- \frac{25}{4} = - \frac{5}{2}

k = \frac{5}{2}, k = - \frac{5}{2}

a^{5} - a^{3} b^{2} - a^{2} b^{3} + b^{5}

\frac{3}{4} - \frac{1}{5}

- x^{2} + 8 x - 7

x^{2} + 3 x < 4

7 b^{2} - 13 b = 10 - 4 b

What are the solutions to the equation k^2-31-2k=-6-3k^2-2k ?
The solutions to the equation k^2-31-2k=-6-3k^2-2k are k= 5/2 ,k=-5/2
Find the zeros of k^2-31-2k=-6-3k^2-2k
The zeros of k^2-31-2k=-6-3k^2-2k are k= 5/2 ,k=-5/2