Given the function f(x)=2x^2 + 3x - 1, find the value(s) of k so that f(k+2) = -2

Answer 1

-2 = f(k+2) = 2(k+2)^2 + 3(k+2) - 1

-2 = 2(k+2)^2 + 3(k+2) - 1

-2 = 2( k^2 + 4k + 4) + 3(k+2) - 1

-2 = 2k^2 + 8k + 8 + 3k + 6 - 1

-2 = 2k^2 + 11k + 13

0 = 2k^2 + 11k + 15

0 = ( 2k + 5 )(k + 3 )

2k + 5 = 0 ---> k = -5/2

k+3=0 ----> k = -3