What are the solutions to the equation (5p-25)(p-7)=1-p ?

The solutions to the equation (5p-25)(p-7)=1-p are p=6,p= 29/5

Find the zeros of (5p-25)(p-7)=1-p

The zeros of (5p-25)(p-7)=1-p are p=6,p= 29/5

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(5 p - 25) (p - 7) = 1 - p

p = 6, p = \frac{29}{5}

Decimal Notation

p = 6, p = 5.8

Solution steps

(5 p - 25) (p - 7) = 1 - p

Expand

(5 p - 25) (p - 7) : 5 p^{2} - 60 p + 175

5 p^{2} - 60 p + 175 = 1 - p

Move

p

to the left side

5 p^{2} - 59 p + 175 = 1

Move

1

to the left side

5 p^{2} - 59 p + 174 = 0

Solve with the quadratic formula

p_{1, 2} = \frac{- ( - 59 ) \pm ( - 59 ) ^{2} - 4 \cdot 5 \cdot 174}{2 \cdot 5}

(- 59)^{2} - 4 \cdot 5 \cdot 174 = 1

p_{1, 2} = \frac{- ( - 59 ) \pm 1}{2 \cdot 5}

Separate the solutions

p_{1} = \frac{- ( - 59 ) + 1}{2 \cdot 5}, p_{2} = \frac{- ( - 59 ) - 1}{2 \cdot 5}

p = \frac{- ( - 59 ) + 1}{2 \cdot 5} : 6

p = \frac{- ( - 59 ) - 1}{2 \cdot 5} : \frac{29}{5}

The solutions to the quadratic equation are:

p = 6, p = \frac{29}{5}

25 x^{2} - 3 x - 12 = - 18 x - 12

1^{2} + 6^{2} = c^{2}

x^{2} - 2 x - 1.25 = 0

(6 x - 9) x = 27

9 x^{2} + αx + 1 = 0

What are the solutions to the equation (5p-25)(p-7)=1-p ?
The solutions to the equation (5p-25)(p-7)=1-p are p=6,p= 29/5
Find the zeros of (5p-25)(p-7)=1-p
The zeros of (5p-25)(p-7)=1-p are p=6,p= 29/5