What are the solutions to the equation k^2+4k-5=0 ?

The solutions to the equation k^2+4k-5=0 are k=1,k=-5

Find the zeros of k^2+4k-5=0

The zeros of k^2+4k-5=0 are k=1,k=-5

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k^{2} + 4 k - 5 = 0

k = 1, k = - 5

Solution steps

k^{2} + 4 k - 5 = 0

Solve with the quadratic formula

k_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 1 \cdot ( - 5 )}{2 \cdot 1}

4^{2} - 4 \cdot 1 \cdot (- 5) = 6

k_{1, 2} = \frac{- 4 \pm 6}{2 \cdot 1}

Separate the solutions

k_{1} = \frac{- 4 + 6}{2 \cdot 1}, k_{2} = \frac{- 4 - 6}{2 \cdot 1}

k = \frac{- 4 + 6}{2 \cdot 1} : 1

k = \frac{- 4 - 6}{2 \cdot 1} : - 5

The solutions to the quadratic equation are:

k = 1, k = - 5

x^{2} + 5 x + 2 = 0

n^{2} + 12 n + 35

\frac{( x - 1 ) ( 3 - x )}{x - 2 ^{2}} \leq 0

- 0.04 X^{2} + 0.64 X + 15 = 0

\frac{6 x ^{2} y ^{3}}{( - 4 x ^{4} y ) ( - 3 x ^{3} y ^{2} )}

What are the solutions to the equation k^2+4k-5=0 ?
The solutions to the equation k^2+4k-5=0 are k=1,k=-5
Find the zeros of k^2+4k-5=0
The zeros of k^2+4k-5=0 are k=1,k=-5