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

The solutions to the equation k^2-k=k^2+6-k^2 are k=3,k=-2

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

The zeros of k^2-k=k^2+6-k^2 are k=3,k=-2

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

k = 3, k = - 2

Solution steps

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

Refine

k^{2} - k = 6

Move

6

to the left side

k^{2} - k - 6 = 0

Solve with the quadratic formula

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

(- 1)^{2} - 4 \cdot 1 \cdot (- 6) = 5

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

Separate the solutions

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

k = \frac{- ( - 1 ) + 5}{2 \cdot 1} : 3

k = \frac{- ( - 1 ) - 5}{2 \cdot 1} : - 2

The solutions to the quadratic equation are:

k = 3, k = - 2

x (x + 1) + 2 = 0

1 8^{2} = x^{2} + x^{2}

x, x (x - 3) = 54

(x + 4)^{2} = 24

8000 = 3 x^{2} + 2 x + 400

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