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

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

Find the zeros of p^2+4p=1

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

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p^{2} + 4 p = 1

p = - 2 + 5, p = - 2 - 5

Solution steps

p^{2} + 4 p = 1

Move

1

to the left side

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

Solve with the quadratic formula

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

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

p_{1, 2} = \frac{- 4 \pm 2 5}{2 \cdot 1}

Separate the solutions

p_{1} = \frac{- 4 + 2 5}{2 \cdot 1}, p_{2} = \frac{- 4 - 2 5}{2 \cdot 1}

p = \frac{- 4 + 2 5}{2 \cdot 1} : - 2 + 5

p = \frac{- 4 - 2 5}{2 \cdot 1} : - 2 - 5

The solutions to the quadratic equation are:

p = - 2 + 5, p = - 2 - 5

[2 v - 6] = [2 v + 4]

(m + 2) 2 - (m + 2) + mn \cdot (n + 2) + 2 n (n + 2)

- 20 + 5 x > 5

x - 1 = - 5

- 2 + 4 x = 6 (x - 4)

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