What are the solutions to the equation 42(p^2-1)=13p ?

The solutions to the equation 42(p^2-1)=13p are p= 7/6 ,p=-6/7

Find the zeros of 42(p^2-1)=13p

The zeros of 42(p^2-1)=13p are p= 7/6 ,p=-6/7

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42 (p^{2} - 1) = 13 p

p = \frac{7}{6}, p = - \frac{6}{7}

Decimal Notation

p = 1.16666 \dots, p = - 0.85714 \dots

Solution steps

42 (p^{2} - 1) = 13 p

Expand

42 (p^{2} - 1) : 42 p^{2} - 42

42 p^{2} - 42 = 13 p

Move

13 p

to the left side

42 p^{2} - 42 - 13 p = 0

Write in the standard form

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

42 p^{2} - 13 p - 42 = 0

Solve with the quadratic formula

p_{1, 2} = \frac{- ( - 13 ) \pm ( - 13 ) ^{2} - 4 \cdot 42 ( - 42 )}{2 \cdot 42}

(- 13)^{2} - 4 \cdot 42 (- 42) = 85

p_{1, 2} = \frac{- ( - 13 ) \pm 85}{2 \cdot 42}

Separate the solutions

p_{1} = \frac{- ( - 13 ) + 85}{2 \cdot 42}, p_{2} = \frac{- ( - 13 ) - 85}{2 \cdot 42}

p = \frac{- ( - 13 ) + 85}{2 \cdot 42} : \frac{7}{6}

p = \frac{- ( - 13 ) - 85}{2 \cdot 42} : - \frac{6}{7}

The solutions to the quadratic equation are:

p = \frac{7}{6}, p = - \frac{6}{7}

x (x + 3) = 180

z^{2} + 14 z - 1 = 0

- 3 x (4 x + 2) = 2 x + 1

4 x^{2} - 72 x = - 68

- 5 x - x^{2} = 84

What are the solutions to the equation 42(p^2-1)=13p ?
The solutions to the equation 42(p^2-1)=13p are p= 7/6 ,p=-6/7
Find the zeros of 42(p^2-1)=13p
The zeros of 42(p^2-1)=13p are p= 7/6 ,p=-6/7