What are the solutions to the equation 6p-4=p^2 ?

The solutions to the equation 6p-4=p^2 are p=3+sqrt(5),p=3-sqrt(5)

Find the zeros of 6p-4=p^2

The zeros of 6p-4=p^2 are p=3+sqrt(5),p=3-sqrt(5)

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6 p - 4 = p^{2}

p = 3 + 5, p = 3 - 5

Decimal Notation

p = 5.23606 \dots, p = 0.76393 \dots

Solution steps

6 p - 4 = p^{2}

Switch sides

p^{2} = 6 p - 4

Move

4

to the left side

p^{2} + 4 = 6 p

Move

6 p

to the left side

p^{2} + 4 - 6 p = 0

Write in the standard form

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

p^{2} - 6 p + 4 = 0

Solve with the quadratic formula

p_{1, 2} = \frac{- ( - 6 ) \pm ( - 6 ) ^{2} - 4 \cdot 1 \cdot 4}{2 \cdot 1}

(- 6)^{2} - 4 \cdot 1 \cdot 4 = 25

p_{1, 2} = \frac{- ( - 6 ) \pm 2 5}{2 \cdot 1}

Separate the solutions

p_{1} = \frac{- ( - 6 ) + 2 5}{2 \cdot 1}, p_{2} = \frac{- ( - 6 ) - 2 5}{2 \cdot 1}

p = \frac{- ( - 6 ) + 2 5}{2 \cdot 1} : 3 + 5

p = \frac{- ( - 6 ) - 2 5}{2 \cdot 1} : 3 - 5

The solutions to the quadratic equation are:

p = 3 + 5, p = 3 - 5

2 x^{4} + 7 x^{3} + 5 x^{2}

\frac{x}{5} - \frac{x}{3} = 2

x (x + 1) > 5

lo g_{3} (2 x + y) - \frac{1}{3} lo g_{3} (x - y)

\frac{x}{2} = \frac{x}{6} - \frac{1}{4}

What are the solutions to the equation 6p-4=p^2 ?
The solutions to the equation 6p-4=p^2 are p=3+sqrt(5),p=3-sqrt(5)
Find the zeros of 6p-4=p^2
The zeros of 6p-4=p^2 are p=3+sqrt(5),p=3-sqrt(5)