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

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

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

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

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

z = 1, z = - 5

Solution steps

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

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

z = 1, z = - 5

(2^{- 3})^{2}

\frac{2 x - 8}{x + 1} \leq 1

- 19 + 8 x = 32 + 5 x

\frac{- 20}{- 2}

3 (d - 8) - 5 = 9 (d + 2) + 1

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