What are the solutions to the equation k^2+12(-2)=k(-2) ?

The solutions to the equation k^2+12(-2)=k(-2) are k=4,k=-6

Find the zeros of k^2+12(-2)=k(-2)

The zeros of k^2+12(-2)=k(-2) are k=4,k=-6

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k^{2} + 12 (- 2) = k (- 2)

k = 4, k = - 6

Solution steps

k^{2} + 12 (- 2) = k (- 2)

Expand

k^{2} + 12 (- 2) : k^{2} - 24

k^{2} - 24 = - 2 k

Move

2 k

to the left side

k^{2} - 24 + 2 k = 0

Write in the standard form

a x^{2} + b x + c = 0

k^{2} + 2 k - 24 = 0

Solve with the quadratic formula

k_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 1 \cdot ( - 24 )}{2 \cdot 1}

2^{2} - 4 \cdot 1 \cdot (- 24) = 10

k_{1, 2} = \frac{- 2 \pm 10}{2 \cdot 1}

Separate the solutions

k_{1} = \frac{- 2 + 10}{2 \cdot 1}, k_{2} = \frac{- 2 - 10}{2 \cdot 1}

k = \frac{- 2 + 10}{2 \cdot 1} : 4

k = \frac{- 2 - 10}{2 \cdot 1} : - 6

The solutions to the quadratic equation are:

k = 4, k = - 6

120 = x^{2} - 1

x, x^{2} - y^{2} + 4 x + 4 = 0

(5 - 2 x)^{2} - 16 = 0

3 b^{2} - 4 b + 1 = 0

x^{2} - 20 x + 40 = - 20

What are the solutions to the equation k^2+12(-2)=k(-2) ?
The solutions to the equation k^2+12(-2)=k(-2) are k=4,k=-6
Find the zeros of k^2+12(-2)=k(-2)
The zeros of k^2+12(-2)=k(-2) are k=4,k=-6