What are the solutions to the equation p^2+49=14p ?

The solutions to the equation p^2+49=14p are p=7

Find the zeros of p^2+49=14p

The zeros of p^2+49=14p are p=7

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p^{2} + 49 = 14 p

p = 7

Solution steps

p^{2} + 49 = 14 p

Move

14 p

to the left side

p^{2} + 49 - 14 p = 0

Write in the standard form

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

p^{2} - 14 p + 49 = 0

Solve with the quadratic formula

p_{1, 2} = \frac{- ( - 14 ) \pm ( - 14 ) ^{2} - 4 \cdot 1 \cdot 49}{2 \cdot 1}

(- 14)^{2} - 4 \cdot 1 \cdot 49 = 0

p_{1, 2} = \frac{- ( - 14 ) \pm 0}{2 \cdot 1}

p = \frac{- ( - 14 )}{2 \cdot 1}

\frac{- ( - 14 )}{2 \cdot 1} = 7

p = 7

The solution to the quadratic equation is:

p = 7

x, f (x - 4) = x^{2} + 2 x - 6

z, 2 q^{2} = z^{2} \cdot p^{2} \cdot (1 - p^{2})

(\frac{( \frac{4}{3} ) a + 5}{3})^{2} = 100

x, (x - 3)^{2} + (y - 6)^{2} = 0

x^{2} + 2 (m - 1) x + (m + 5) = 0

What are the solutions to the equation p^2+49=14p ?
The solutions to the equation p^2+49=14p are p=7
Find the zeros of p^2+49=14p
The zeros of p^2+49=14p are p=7