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

The solutions to the equation k^2+2k=3 are k=1,k=-3

Find the zeros of k^2+2k=3

The zeros of k^2+2k=3 are k=1,k=-3

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k^{2} + 2 k = 3

k = 1, k = - 3

Solution steps

k^{2} + 2 k = 3

Move

3

to the left side

k^{2} + 2 k - 3 = 0

Solve with the quadratic formula

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

2^{2} - 4 \cdot 1 \cdot (- 3) = 4

k_{1, 2} = \frac{- 2 \pm 4}{2 \cdot 1}

Separate the solutions

k_{1} = \frac{- 2 + 4}{2 \cdot 1}, k_{2} = \frac{- 2 - 4}{2 \cdot 1}

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

k = \frac{- 2 - 4}{2 \cdot 1} : - 3

The solutions to the quadratic equation are:

k = 1, k = - 3

3 y + 16 = 5 y

\frac{2 π}{\frac{2 π}{7}}

d + 3 > - 5

- 23 \leq 4 x - 3 \leq 1

16 x - 5 = 18 x - 2

What are the solutions to the equation k^2+2k=3 ?
The solutions to the equation k^2+2k=3 are k=1,k=-3
Find the zeros of k^2+2k=3
The zeros of k^2+2k=3 are k=1,k=-3