What are the solutions to the equation r^2+2r-7=0 ?

The solutions to the equation r^2+2r-7=0 are r=-1+2sqrt(2),r=-1-2sqrt(2)

Find the zeros of r^2+2r-7=0

The zeros of r^2+2r-7=0 are r=-1+2sqrt(2),r=-1-2sqrt(2)

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r^{2} + 2 r - 7 = 0

r = - 1 + 22, r = - 1 - 22

Decimal Notation

r = 1.82842 \dots, r = - 3.82842 \dots

Solution steps

r^{2} + 2 r - 7 = 0

Solve with the quadratic formula

r_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 1 \cdot ( - 7 )}{2 \cdot 1}

2^{2} - 4 \cdot 1 \cdot (- 7) = 42

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

Separate the solutions

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

r = \frac{- 2 + 4 2}{2 \cdot 1} : - 1 + 22

r = \frac{- 2 - 4 2}{2 \cdot 1} : - 1 - 22

The solutions to the quadratic equation are:

r = - 1 + 22, r = - 1 - 22

(u)^{2} - 17 u + 16 = 0

x, x^{2} - k x + 9 = 0

- 8 y^{2} = - 16

x^{2} + 6 x - 8 = 0

48 = x^{2}

What are the solutions to the equation r^2+2r-7=0 ?
The solutions to the equation r^2+2r-7=0 are r=-1+2sqrt(2),r=-1-2sqrt(2)
Find the zeros of r^2+2r-7=0
The zeros of r^2+2r-7=0 are r=-1+2sqrt(2),r=-1-2sqrt(2)