What are the solutions to the equation 12(p^2-1)=7p ?

The solutions to the equation 12(p^2-1)=7p are p= 4/3 ,p=-3/4

Find the zeros of 12(p^2-1)=7p

The zeros of 12(p^2-1)=7p are p= 4/3 ,p=-3/4

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12 (p^{2} - 1) = 7 p

p = \frac{4}{3}, p = - \frac{3}{4}

Decimal Notation

p = 1.33333 \dots, p = - 0.75

Solution steps

12 (p^{2} - 1) = 7 p

Expand

12 (p^{2} - 1) : 12 p^{2} - 12

12 p^{2} - 12 = 7 p

Move

7 p

to the left side

12 p^{2} - 12 - 7 p = 0

Write in the standard form

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

12 p^{2} - 7 p - 12 = 0

Solve with the quadratic formula

p_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 \cdot 12 ( - 12 )}{2 \cdot 12}

(- 7)^{2} - 4 \cdot 12 (- 12) = 25

p_{1, 2} = \frac{- ( - 7 ) \pm 25}{2 \cdot 12}

Separate the solutions

p_{1} = \frac{- ( - 7 ) + 25}{2 \cdot 12}, p_{2} = \frac{- ( - 7 ) - 25}{2 \cdot 12}

p = \frac{- ( - 7 ) + 25}{2 \cdot 12} : \frac{4}{3}

p = \frac{- ( - 7 ) - 25}{2 \cdot 12} : - \frac{3}{4}

The solutions to the quadratic equation are:

p = \frac{4}{3}, p = - \frac{3}{4}

6^{- 7}

\frac{- 10}{5}

x + 5 = 2 x - 7

x^{2} < 36

10 m - 3 > 6 m + 1

What are the solutions to the equation 12(p^2-1)=7p ?
The solutions to the equation 12(p^2-1)=7p are p= 4/3 ,p=-3/4
Find the zeros of 12(p^2-1)=7p
The zeros of 12(p^2-1)=7p are p= 4/3 ,p=-3/4