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

The solutions to the equation r^2+4r+2=0 are r=-2+sqrt(2),r=-2-sqrt(2)

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

The zeros of r^2+4r+2=0 are r=-2+sqrt(2),r=-2-sqrt(2)

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

r = - 2 + 2, r = - 2 - 2

Solution steps

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

Solve with the quadratic formula

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

4^{2} - 4 \cdot 1 \cdot 2 = 22

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

r = - 2 + 2, r = - 2 - 2

x (x + 5) = 0

- 25 - \frac{x}{4} = - 12

2 x^{2} - 1 = 11 x

6 4^{\frac{- 1}{6}}

y^{3} - 8 x^{3}

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