Z^2 - 6z - 27 = 0 you need to factor out firts then solve for z
(z+3 )(z-9) = 0
z+ 3 = 0
z = -3
z-9= 0
z = 9
I have no idea what the answer is
{k^2 + 2k + 1 = 25, k^2 + 2k + 1 = 64}
{k^2 + 2k - 24 = 0, k^2 + 2k - 63 = 0}
{(k - 4)(k + 6) = 0, (k - 7)(k + 9) = 0}
{k = -6, 4, -9, 7)
Therefore, range = {-9, -6, 4, 7)
I see the picture but its really bad so 5.-2
Check if you can write an equation relating the term number to the actual value
n1=3
n2=10 = 3+7
n3= 17 = n2+7 = n1+7+7 = n1 +2*7
n4= 24 = n1+3*7
so you will notice a pattern
for the x-th term
n_x =3+(x-1)*7
the 50th term would be n_50 = 3+(50-1) * 7
