What are the solutions to the equation 40(p^2-1)=39p ?

The solutions to the equation 40(p^2-1)=39p are p= 8/5 ,p=-5/8

Find the zeros of 40(p^2-1)=39p

The zeros of 40(p^2-1)=39p are p= 8/5 ,p=-5/8

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40 (p^{2} - 1) = 39 p

p = \frac{8}{5}, p = - \frac{5}{8}

Decimal Notation

p = 1.6, p = - 0.625

Solution steps

40 (p^{2} - 1) = 39 p

Expand

40 (p^{2} - 1) : 40 p^{2} - 40

40 p^{2} - 40 = 39 p

Move

39 p

to the left side

40 p^{2} - 40 - 39 p = 0

Write in the standard form

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

40 p^{2} - 39 p - 40 = 0

Solve with the quadratic formula

p_{1, 2} = \frac{- ( - 39 ) \pm ( - 39 ) ^{2} - 4 \cdot 40 ( - 40 )}{2 \cdot 40}

(- 39)^{2} - 4 \cdot 40 (- 40) = 89

p_{1, 2} = \frac{- ( - 39 ) \pm 89}{2 \cdot 40}

Separate the solutions

p_{1} = \frac{- ( - 39 ) + 89}{2 \cdot 40}, p_{2} = \frac{- ( - 39 ) - 89}{2 \cdot 40}

p = \frac{- ( - 39 ) + 89}{2 \cdot 40} : \frac{8}{5}

p = \frac{- ( - 39 ) - 89}{2 \cdot 40} : - \frac{5}{8}

The solutions to the quadratic equation are:

p = \frac{8}{5}, p = - \frac{5}{8}

4 x^{2} - 23 x + 13 = - 3 x + 4

5 z^{2} - 21 z + 11 = - 4 z

2 x^{2} + 11 x - 63 = 0

a^{2} + 4 = 4 a

3 x^{2} - 12 x = - 8

What are the solutions to the equation 40(p^2-1)=39p ?
The solutions to the equation 40(p^2-1)=39p are p= 8/5 ,p=-5/8
Find the zeros of 40(p^2-1)=39p
The zeros of 40(p^2-1)=39p are p= 8/5 ,p=-5/8