What are the solutions to the equation 5p^2=-10p+1 ?

The solutions to the equation 5p^2=-10p+1 are p=(-5+sqrt(30))/5 ,p=-(5+sqrt(30))/5

Find the zeros of 5p^2=-10p+1

The zeros of 5p^2=-10p+1 are p=(-5+sqrt(30))/5 ,p=-(5+sqrt(30))/5

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5 p^{2} = - 10 p + 1

p = \frac{- 5 + 30}{5}, p = - \frac{5 + 30}{5}

Solution steps

5 p^{2} = - 10 p + 1

Move

1

to the left side

5 p^{2} - 1 = - 10 p

Move

10 p

to the left side

5 p^{2} - 1 + 10 p = 0

Write in the standard form

a x^{2} + b x + c = 0

5 p^{2} + 10 p - 1 = 0

Solve with the quadratic formula

p_{1, 2} = \frac{- 10 \pm 1 0 ^{2} - 4 \cdot 5 ( - 1 )}{2 \cdot 5}

1 0^{2} - 4 \cdot 5 (- 1) = 230

p_{1, 2} = \frac{- 10 \pm 2 30}{2 \cdot 5}

Separate the solutions

p_{1} = \frac{- 10 + 2 30}{2 \cdot 5}, p_{2} = \frac{- 10 - 2 30}{2 \cdot 5}

p = \frac{- 10 + 2 30}{2 \cdot 5} : \frac{- 5 + 30}{5}

p = \frac{- 10 - 2 30}{2 \cdot 5} : - \frac{5 + 30}{5}

The solutions to the quadratic equation are:

p = \frac{- 5 + 30}{5}, p = - \frac{5 + 30}{5}

x + 9 = 15

10 x^{2} - 8 x - 4 = 0

\frac{2 ^{3} + 2 ^{- 3}}{4 ^{3} + 1}

∣ 6 x + 5 ∣ = 8

∣ k + 1 ∣ + 2 = 8

What are the solutions to the equation 5p^2=-10p+1 ?
The solutions to the equation 5p^2=-10p+1 are p=(-5+sqrt(30))/5 ,p=-(5+sqrt(30))/5
Find the zeros of 5p^2=-10p+1
The zeros of 5p^2=-10p+1 are p=(-5+sqrt(30))/5 ,p=-(5+sqrt(30))/5