What are the solutions to the equation 7p^2=35p+168 ?

The solutions to the equation 7p^2=35p+168 are p=8,p=-3

Find the zeros of 7p^2=35p+168

The zeros of 7p^2=35p+168 are p=8,p=-3

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7 p^{2} = 35 p + 168

p = 8, p = - 3

Solution steps

7 p^{2} = 35 p + 168

Move

168

to the left side

7 p^{2} - 168 = 35 p

Move

35 p

to the left side

7 p^{2} - 168 - 35 p = 0

Write in the standard form

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

7 p^{2} - 35 p - 168 = 0

Solve with the quadratic formula

p_{1, 2} = \frac{- ( - 35 ) \pm ( - 35 ) ^{2} - 4 \cdot 7 ( - 168 )}{2 \cdot 7}

(- 35)^{2} - 4 \cdot 7 (- 168) = 77

p_{1, 2} = \frac{- ( - 35 ) \pm 77}{2 \cdot 7}

Separate the solutions

p_{1} = \frac{- ( - 35 ) + 77}{2 \cdot 7}, p_{2} = \frac{- ( - 35 ) - 77}{2 \cdot 7}

p = \frac{- ( - 35 ) + 77}{2 \cdot 7} : 8

p = \frac{- ( - 35 ) - 77}{2 \cdot 7} : - 3

The solutions to the quadratic equation are:

p = 8, p = - 3

9 x^{2} - 48 x + 54 = 0

4.905 t^{2} - 90.27 t - 150 = 0

3 x^{2} + 30 x + 72 = 0

4 x^{2} + 43 x + 3 = 0

x, y = x^{2} + p x + q

What are the solutions to the equation 7p^2=35p+168 ?
The solutions to the equation 7p^2=35p+168 are p=8,p=-3
Find the zeros of 7p^2=35p+168
The zeros of 7p^2=35p+168 are p=8,p=-3