What are the solutions to the equation p(p+1)=2 ?

The solutions to the equation p(p+1)=2 are p=1,p=-2

Find the zeros of p(p+1)=2

The zeros of p(p+1)=2 are p=1,p=-2

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p (p + 1) = 2

p = 1, p = - 2

Solution steps

p (p + 1) = 2

Expand

p (p + 1) : p^{2} + p

p^{2} + p = 2

Move

2

to the left side

p^{2} + p - 2 = 0

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

p = \frac{- 1 - 3}{2 \cdot 1} : - 2

The solutions to the quadratic equation are:

p = 1, p = - 2

(- 3)^{- 1}

x^{\frac{2}{3}} \cdot x^{\frac{11}{3}}

- 8 + \frac{16}{3}

4 m^{2} - 4 m + 1 = 0

6 x^{2} + 2 x = 4

What are the solutions to the equation p(p+1)=2 ?
The solutions to the equation p(p+1)=2 are p=1,p=-2
Find the zeros of p(p+1)=2
The zeros of p(p+1)=2 are p=1,p=-2