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

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

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

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

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

n = 1, n = - 5

Solution steps

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

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

n = 1, n = - 5

4 + \frac{x}{2} - \frac{x ^{2}}{8} = 0

3 x^{2} - 16 x - 35 = 0

(k - 1) x^{2} + 2 x + (2 k - 3) = 0

a = π \cdot r^{2}

\frac{1}{4} r^{2} + 3 r + 8 = 0

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