What are the solutions to the equation k^2-2k-7=0 ?

The solutions to the equation k^2-2k-7=0 are k=1+2sqrt(2),k=1-2sqrt(2)

Find the zeros of k^2-2k-7=0

The zeros of k^2-2k-7=0 are k=1+2sqrt(2),k=1-2sqrt(2)

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k^{2} - 2 k - 7 = 0

k = 1 + 22, k = 1 - 22

Decimal Notation

k = 3.82842 \dots, k = - 1.82842 \dots

Solution steps

k^{2} - 2 k - 7 = 0

Solve with the quadratic formula

k_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 1 \cdot ( - 7 )}{2 \cdot 1}

(- 2)^{2} - 4 \cdot 1 \cdot (- 7) = 42

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

Separate the solutions

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

k = \frac{- ( - 2 ) + 4 2}{2 \cdot 1} : 1 + 22

k = \frac{- ( - 2 ) - 4 2}{2 \cdot 1} : 1 - 22

The solutions to the quadratic equation are:

k = 1 + 22, k = 1 - 22

18 x^{2} + 6 x + 1 = 0

z^{2} - 11 z + 24 = 0

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

3 x (x - 2) - (x - 6) = 23 (x - 6)

4 x^{2} + 2 = - 5 x - 5 x

What are the solutions to the equation k^2-2k-7=0 ?
The solutions to the equation k^2-2k-7=0 are k=1+2sqrt(2),k=1-2sqrt(2)
Find the zeros of k^2-2k-7=0
The zeros of k^2-2k-7=0 are k=1+2sqrt(2),k=1-2sqrt(2)