What are the solutions to the equation 5(p^2-4)-1=4 ?

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

Find the zeros of 5(p^2-4)-1=4

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

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5 (p^{2} - 4) - 1 = 4

p = 5, p = - 5

Decimal Notation

p = 2.23606 \dots, p = - 2.23606 \dots

Solution steps

5 (p^{2} - 4) - 1 = 4

Move

1

to the right side

5 (p^{2} - 4) = 5

Divide both sides by

5

p^{2} - 4 = 1

Move

4

to the right side

p^{2} = 5

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

p = 5, p = - 5

- 4 x^{2} + 8 x = 7

3 - 2 x^{2} = - 5

\frac{7}{5} x^{2} = 2

y^{2} - 12 y = - 25

x^{2} - 35 x = 60

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