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

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

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

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

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

x = 1, x = - 5

Solution steps

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

Switch sides

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

12 - 2 x = 8 x - 18

(\frac{3}{2}) 3

6^{2} - 4 \cdot 2 \cdot 3

216 k^{4}

- x^{2} - x + 12 = 0

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