Question 946196

{{{f(x)=2x^2+3x-1}}}, 

the value(s) of {{{k}}} so that {{{f(k+2)=-2}}} will be:

{{{-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+13+2}}}

{{{2k^2+11k+15=0}}}

{{{2k^2+5k+6k+15 = 0}}}

{{{(2k^2+5k)+(6k+15) = 0}}}

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

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

solutions:

if {{{(k+3) = 0}}}=> {{{k=-3}}}

if {{{(2k+5) = 0}}}=>{{{2k=-5}}}=>{{{k=-5/2}}}