What are the solutions to the equation p^2-5p=-4 ?

The solutions to the equation p^2-5p=-4 are p=4,p=1

Find the zeros of p^2-5p=-4

The zeros of p^2-5p=-4 are p=4,p=1

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p^{2} - 5 p = - 4

p = 4, p = 1

Solution steps

p^{2} - 5 p = - 4

Move

4

to the left side

p^{2} - 5 p + 4 = 0

Solve with the quadratic formula

p_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 1 \cdot 4}{2 \cdot 1}

(- 5)^{2} - 4 \cdot 1 \cdot 4 = 3

p_{1, 2} = \frac{- ( - 5 ) \pm 3}{2 \cdot 1}

Separate the solutions

p_{1} = \frac{- ( - 5 ) + 3}{2 \cdot 1}, p_{2} = \frac{- ( - 5 ) - 3}{2 \cdot 1}

p = \frac{- ( - 5 ) + 3}{2 \cdot 1} : 4

p = \frac{- ( - 5 ) - 3}{2 \cdot 1} : 1

The solutions to the quadratic equation are:

p = 4, p = 1

20 x + 7 = 9 x + 95

\frac{1}{10} + \frac{3}{5}

1331 x^{3} - 8 y^{3}

(8 a^{- 9})^{- \frac{2}{3}}

5 + 6 x = 7

What are the solutions to the equation p^2-5p=-4 ?
The solutions to the equation p^2-5p=-4 are p=4,p=1
Find the zeros of p^2-5p=-4
The zeros of p^2-5p=-4 are p=4,p=1