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

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

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

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

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

r = - 1 + 5, r = - 1 - 5

Decimal Notation

r = 1.23606 \dots, r = - 3.23606 \dots

Solution steps

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

Solve with the quadratic formula

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

2^{2} - 4 \cdot 1 \cdot (- 4) = 25

r_{1, 2} = \frac{- 2 \pm 2 5}{2 \cdot 1}

Separate the solutions

r_{1} = \frac{- 2 + 2 5}{2 \cdot 1}, r_{2} = \frac{- 2 - 2 5}{2 \cdot 1}

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

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

The solutions to the quadratic equation are:

r = - 1 + 5, r = - 1 - 5

(x - 2)^{2} > 4

48 x^{2} - 240 x + 57

14 c^{2} + 2 c

- 4.9 x^{2} + 42.4 x - 20 = 0

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