What are the solutions to the equation 10(p^2-1)=21p ?

The solutions to the equation 10(p^2-1)=21p are p= 5/2 ,p=-2/5

Find the zeros of 10(p^2-1)=21p

The zeros of 10(p^2-1)=21p are p= 5/2 ,p=-2/5

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10 (p^{2} - 1) = 21 p

p = \frac{5}{2}, p = - \frac{2}{5}

Decimal

p = 2.5, p = - 0.4

Solution steps

10 (p^{2} - 1) = 21 p

Expand

10 (p^{2} - 1) : 10 p^{2} - 10

10 p^{2} - 10 = 21 p

Move

21 p

to the left side

10 p^{2} - 10 - 21 p = 0

Write in the standard form

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

10 p^{2} - 21 p - 10 = 0

Solve with the quadratic formula

p_{1, 2} = \frac{- ( - 21 ) \pm ( - 21 ) ^{2} - 4 \cdot 10 ( - 10 )}{2 \cdot 10}

(- 21)^{2} - 4 \cdot 10 (- 10) = 29

p_{1, 2} = \frac{- ( - 21 ) \pm 29}{2 \cdot 10}

Separate the solutions

p_{1} = \frac{- ( - 21 ) + 29}{2 \cdot 10}, p_{2} = \frac{- ( - 21 ) - 29}{2 \cdot 10}

p = \frac{- ( - 21 ) + 29}{2 \cdot 10} : \frac{5}{2}

p = \frac{- ( - 21 ) - 29}{2 \cdot 10} : - \frac{2}{5}

The solutions to the quadratic equation are:

p = \frac{5}{2}, p = - \frac{2}{5}

- 16 t^{2} + 70 t + 1 = 41

3 x^{2} + 14 - 5 = 0

- 5 x^{2} + 80 = 0

3 x^{2} = 6 x - 12

x, 2 x^{2} y^{2} = c

What are the solutions to the equation 10(p^2-1)=21p ?
The solutions to the equation 10(p^2-1)=21p are p= 5/2 ,p=-2/5
Find the zeros of 10(p^2-1)=21p
The zeros of 10(p^2-1)=21p are p= 5/2 ,p=-2/5