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

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

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

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

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

x = 1, x = - 5

Solution steps

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

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = 1, x = - 5

4 r^{2} + 44 r + 121 = 0

9 (x + 2)^{2} - 4 = 2

- 12 (x - 8) = 3 (x + 17)

125 - 300 n + 240 n^{2} - 64 n^{3}

30 a^{2} - 57 a - 45

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