What are the solutions to the equation 6(p^2-1)=5p ?

The solutions to the equation 6(p^2-1)=5p are p= 3/2 ,p=-2/3

Find the zeros of 6(p^2-1)=5p

The zeros of 6(p^2-1)=5p are p= 3/2 ,p=-2/3

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

p = \frac{3}{2}, p = - \frac{2}{3}

Decimal Notation

p = 1.5, p = - 0.66666 \dots

Solution steps

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

Expand

6 (p^{2} - 1) : 6 p^{2} - 6

6 p^{2} - 6 = 5 p

Move

5 p

to the left side

6 p^{2} - 6 - 5 p = 0

Write in the standard form

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

6 p^{2} - 5 p - 6 = 0

Solve with the quadratic formula

p_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 6 ( - 6 )}{2 \cdot 6}

(- 5)^{2} - 4 \cdot 6 (- 6) = 13

p_{1, 2} = \frac{- ( - 5 ) \pm 13}{2 \cdot 6}

Separate the solutions

p_{1} = \frac{- ( - 5 ) + 13}{2 \cdot 6}, p_{2} = \frac{- ( - 5 ) - 13}{2 \cdot 6}

p = \frac{- ( - 5 ) + 13}{2 \cdot 6} : \frac{3}{2}

p = \frac{- ( - 5 ) - 13}{2 \cdot 6} : - \frac{2}{3}

The solutions to the quadratic equation are:

p = \frac{3}{2}, p = - \frac{2}{3}

w^{2} - 9

(\frac{6}{31} - \frac{12}{31}) \cdot (12)

- 9 a - 1 \leq 26

2 (3 x - 1) - 5 (x - 2) < 3 (x - 2)

x^{2} + 0^{2} - 14 x + 4 (0) + 28 = 0

What are the solutions to the equation 6(p^2-1)=5p ?
The solutions to the equation 6(p^2-1)=5p are p= 3/2 ,p=-2/3
Find the zeros of 6(p^2-1)=5p
The zeros of 6(p^2-1)=5p are p= 3/2 ,p=-2/3