What are the solutions to the equation r^2+2r-8=0 ?

The solutions to the equation r^2+2r-8=0 are r=2,r=-4

Find the zeros of r^2+2r-8=0

The zeros of r^2+2r-8=0 are r=2,r=-4

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r^{2} + 2 r - 8 = 0

r = 2, r = - 4

Solution steps

r^{2} + 2 r - 8 = 0

Solve with the quadratic formula

r_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 1 \cdot ( - 8 )}{2 \cdot 1}

2^{2} - 4 \cdot 1 \cdot (- 8) = 6

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

r = 2, r = - 4

- 5 x < 25

2 0^{- 8} \cdot 2 0^{8}

x^{2} - 21 = 4 x

3 x^{- 4} y^{9} \cdot 8 x^{- 1} y^{- 6}

12 8^{\frac{1}{4}} \cdot 2^{\frac{1}{4}}

What are the solutions to the equation r^2+2r-8=0 ?
The solutions to the equation r^2+2r-8=0 are r=2,r=-4
Find the zeros of r^2+2r-8=0
The zeros of r^2+2r-8=0 are r=2,r=-4