What are the solutions to the equation r^2+3r=1 ?

The solutions to the equation r^2+3r=1 are r=(-3+sqrt(13))/2 ,r=(-3-sqrt(13))/2

Find the zeros of r^2+3r=1

The zeros of r^2+3r=1 are r=(-3+sqrt(13))/2 ,r=(-3-sqrt(13))/2

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r^{2} + 3 r = 1

r = \frac{- 3 + 13}{2}, r = \frac{- 3 - 13}{2}

Decimal Notation

r = 0.30277 \dots, r = - 3.30277 \dots

Solution steps

r^{2} + 3 r = 1

Move

1

to the left side

r^{2} + 3 r - 1 = 0

Solve with the quadratic formula

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

3^{2} - 4 \cdot 1 \cdot (- 1) = 13

r_{1, 2} = \frac{- 3 \pm 13}{2 \cdot 1}

Separate the solutions

r_{1} = \frac{- 3 + 13}{2 \cdot 1}, r_{2} = \frac{- 3 - 13}{2 \cdot 1}

r = \frac{- 3 + 13}{2 \cdot 1} : \frac{- 3 + 13}{2}

r = \frac{- 3 - 13}{2 \cdot 1} : \frac{- 3 - 13}{2}

The solutions to the quadratic equation are:

r = \frac{- 3 + 13}{2}, r = \frac{- 3 - 13}{2}

- 2 x^{2} + 4 x

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

- 6 x^{2} + 6 x + 12 = 0

x^{2} + 15 x + 16 = 0

x, x^{2} + 2 x - 15 = 0

What are the solutions to the equation r^2+3r=1 ?
The solutions to the equation r^2+3r=1 are r=(-3+sqrt(13))/2 ,r=(-3-sqrt(13))/2
Find the zeros of r^2+3r=1
The zeros of r^2+3r=1 are r=(-3+sqrt(13))/2 ,r=(-3-sqrt(13))/2