What are the solutions to the equation p^2-4p=21 ?

The solutions to the equation p^2-4p=21 are p=7,p=-3

Find the zeros of p^2-4p=21

The zeros of p^2-4p=21 are p=7,p=-3

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p^{2} - 4 p = 21

p = 7, p = - 3

Solution steps

p^{2} - 4 p = 21

Move

21

to the left side

p^{2} - 4 p - 21 = 0

Solve with the quadratic formula

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

(- 4)^{2} - 4 \cdot 1 \cdot (- 21) = 10

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

Separate the solutions

p_{1} = \frac{- ( - 4 ) + 10}{2 \cdot 1}, p_{2} = \frac{- ( - 4 ) - 10}{2 \cdot 1}

p = \frac{- ( - 4 ) + 10}{2 \cdot 1} : 7

p = \frac{- ( - 4 ) - 10}{2 \cdot 1} : - 3

The solutions to the quadratic equation are:

p = 7, p = - 3

∣ 4 x - 7 ∣ = 8

a + \frac{1}{b + \frac{1}{c + \frac{1}{d}}}

x + 8 = - 2

\frac{3}{2 - 7}

\frac{5.6}{8} \cdot (x + 8) = \frac{6}{12} \cdot (x + 12)

What are the solutions to the equation p^2-4p=21 ?
The solutions to the equation p^2-4p=21 are p=7,p=-3
Find the zeros of p^2-4p=21
The zeros of p^2-4p=21 are p=7,p=-3