Question

How do you solve `p( p + 5) = 14`?

Answer 1

`p=2,p=-7`

Explanation:

`p(p+5)=14`

# :.p^2+5p=14#

`:.p^2+5p-14=0`

`:.(p-2)(p+7)=0`

`:.p=2` or `p=-7`

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check;-

substitute `p=2`

`:.2(2+5)=14`

`:.14=14`

substitute `p=-7`

`;.-7(-7+5)=14`

`:.-7(-2)=14`

`:.14=14`