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

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

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

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

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

r = 1, r = - 5

Solution steps

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

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

r = 1, r = - 5

- \frac{10}{8}

5 v + 3 = 5 v - 5

x \cdot x^{- 1}

- 14 (z - 5) = - 14 z + 70

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