What are the solutions to the equation r^2+4r=-1 ?

The solutions to the equation r^2+4r=-1 are r=-2+sqrt(3),r=-2-sqrt(3)

Find the zeros of r^2+4r=-1

The zeros of r^2+4r=-1 are r=-2+sqrt(3),r=-2-sqrt(3)

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r^{2} + 4 r = - 1

r = - 2 + 3, r = - 2 - 3

Decimal Notation

r = - 0.26794 \dots, r = - 3.73205 \dots

Solution steps

r^{2} + 4 r = - 1

Move

1

to the left side

r^{2} + 4 r + 1 = 0

Solve with the quadratic formula

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

4^{2} - 4 \cdot 1 \cdot 1 = 23

r_{1, 2} = \frac{- 4 \pm 2 3}{2 \cdot 1}

Separate the solutions

r_{1} = \frac{- 4 + 2 3}{2 \cdot 1}, r_{2} = \frac{- 4 - 2 3}{2 \cdot 1}

r = \frac{- 4 + 2 3}{2 \cdot 1} : - 2 + 3

r = \frac{- 4 - 2 3}{2 \cdot 1} : - 2 - 3

The solutions to the quadratic equation are:

r = - 2 + 3, r = - 2 - 3

7 x^{2} - 2 x = 5

7 x^{2} - 2 x = 2

3 x^{2} = 2 (x - 1)^{2} + 3

x^{(2)} + 20 x + 28 = 9

6 - 5 x + x^{2} = 0

What are the solutions to the equation r^2+4r=-1 ?
The solutions to the equation r^2+4r=-1 are r=-2+sqrt(3),r=-2-sqrt(3)
Find the zeros of r^2+4r=-1
The zeros of r^2+4r=-1 are r=-2+sqrt(3),r=-2-sqrt(3)