What are the solutions to the equation 56(p^2-1)=15p ?

The solutions to the equation 56(p^2-1)=15p are p= 8/7 ,p=-7/8

Find the zeros of 56(p^2-1)=15p

The zeros of 56(p^2-1)=15p are p= 8/7 ,p=-7/8

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56 (p^{2} - 1) = 15 p

p = \frac{8}{7}, p = - \frac{7}{8}

Decimal Notation

p = 1.14285 \dots, p = - 0.875

Solution steps

56 (p^{2} - 1) = 15 p

Expand

56 (p^{2} - 1) : 56 p^{2} - 56

56 p^{2} - 56 = 15 p

Move

15 p

to the left side

56 p^{2} - 56 - 15 p = 0

Write in the standard form

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

56 p^{2} - 15 p - 56 = 0

Solve with the quadratic formula

p_{1, 2} = \frac{- ( - 15 ) \pm ( - 15 ) ^{2} - 4 \cdot 56 ( - 56 )}{2 \cdot 56}

(- 15)^{2} - 4 \cdot 56 (- 56) = 113

p_{1, 2} = \frac{- ( - 15 ) \pm 113}{2 \cdot 56}

Separate the solutions

p_{1} = \frac{- ( - 15 ) + 113}{2 \cdot 56}, p_{2} = \frac{- ( - 15 ) - 113}{2 \cdot 56}

p = \frac{- ( - 15 ) + 113}{2 \cdot 56} : \frac{8}{7}

p = \frac{- ( - 15 ) - 113}{2 \cdot 56} : - \frac{7}{8}

The solutions to the quadratic equation are:

p = \frac{8}{7}, p = - \frac{7}{8}

w^{2} - 4 w - 3 = 0

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

- 3 (t - 1)^{2} + 12 = 0

- x^{2} + 9 = 25

x^{2} + 23 - 140 = 0

What are the solutions to the equation 56(p^2-1)=15p ?
The solutions to the equation 56(p^2-1)=15p are p= 8/7 ,p=-7/8
Find the zeros of 56(p^2-1)=15p
The zeros of 56(p^2-1)=15p are p= 8/7 ,p=-7/8