What are the solutions to the equation k^2-2k-3=0 ?

The solutions to the equation k^2-2k-3=0 are k=3,k=-1

Find the zeros of k^2-2k-3=0

The zeros of k^2-2k-3=0 are k=3,k=-1

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

k = 3, k = - 1

Solution steps

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} : 3

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

The solutions to the quadratic equation are:

k = 3, k = - 1

6 x^{2} - 12 x + 1476 = 0

x, x + 2 x^{2} - 4 = - 7 x + 6

3 x + 2 x^{2} = 7

x^{2} - 2.454 x + 1.5 = 0

8 x^{2} + 13 x + 4 = 3 x^{2}

What are the solutions to the equation k^2-2k-3=0 ?
The solutions to the equation k^2-2k-3=0 are k=3,k=-1
Find the zeros of k^2-2k-3=0
The zeros of k^2-2k-3=0 are k=3,k=-1