Question

find the number of sides of a polygon having 35 diagonals

Answer 1

by the formula
 n(n-3)/2 = no. of diagonals

n(n-3) = 35 *2                            (as no. of diagonals are 35)

n^2 - 3n = 70

n^2 - 3n -70 =0                (factorizing it)

(n+7)(n-10)=0

n-10 = 0

so n=10 .
no. of sides are 10.
verifying it
n(n-3)/2 =no. of diagonals
10(10-3)/2 =no.of diagonals
10*7 /2  = no. of diagonals
35 =no. of diagonals

Answer 2

n(n-3)/2=35
=>n*n-3n=70
=>n*n-3n-70=0
=>n*n-10n+7n-70=0
=>n(n-10)+7(n-10)=0
=>n=10(since -7 isn't possible)