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

Decimal Notation

p = 4.23606 \dots, p = - 0.23606 \dots

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

x, w = x^{2} + y^{2} + z^{2}

- 4 r^{2} - 4 r = 6

36 x^{2} + 84 x + 49 = 0

x^{2} = 1 0^{2} + 6^{2}

x^{2} - 2 x - 5

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)