What are the solutions to the equation -k^2+20k+12=0 ?

The solutions to the equation -k^2+20k+12=0 are k=-2(2sqrt(7)-5),k=2(5+2sqrt(7))

Find the zeros of -k^2+20k+12=0

The zeros of -k^2+20k+12=0 are k=-2(2sqrt(7)-5),k=2(5+2sqrt(7))

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- k^{2} + 20 k + 12 = 0

k = - 2 (27 - 5), k = 2 (5 + 27)

Decimal Notation

k = - 0.58300 \dots, k = 20.58300 \dots

Solution steps

- k^{2} + 20 k + 12 = 0

Solve with the quadratic formula

k_{1, 2} = \frac{- 20 \pm 2 0 ^{2} - 4 ( - 1 ) \cdot 12}{2 ( - 1 )}

2 0^{2} - 4 (- 1) \cdot 12 = 87

k_{1, 2} = \frac{- 20 \pm 8 7}{2 ( - 1 )}

Separate the solutions

k_{1} = \frac{- 20 + 8 7}{2 ( - 1 )}, k_{2} = \frac{- 20 - 8 7}{2 ( - 1 )}

k = \frac{- 20 + 8 7}{2 ( - 1 )} : - 2 (27 - 5)

k = \frac{- 20 - 8 7}{2 ( - 1 )} : 2 (5 + 27)

The solutions to the quadratic equation are:

k = - 2 (27 - 5), k = 2 (5 + 27)

6^{2} = a^{2} - 8^{2}

x^{2} + (- 3) x + (- 3) + 8 = 0

x + 6 = x^{2} - 2

- x^{2} + 20 x - 36 = 0

a^{2} - 2 a - 2 = 0

What are the solutions to the equation -k^2+20k+12=0 ?
The solutions to the equation -k^2+20k+12=0 are k=-2(2sqrt(7)-5),k=2(5+2sqrt(7))
Find the zeros of -k^2+20k+12=0
The zeros of -k^2+20k+12=0 are k=-2(2sqrt(7)-5),k=2(5+2sqrt(7))