It is given that 24x2−22x2+x+16+22x+30=0, which can be written as: ‌=>(22x2)2−22x2⋅2x+15⋅21+(2x+15)2=0 ‌=>(22x2−2x+15)2=0 ‌=22x2−2x+15=0 ( Since (a−b)2=0=>a−b=0) ‌=>2x2=x+15 ‌=2x2−x−15=0 ‌=>2x2−6x+5x−15=0 ‌⇒2x(x−3)+5(x−3)=0 ‌⇒(2x+5)(x−3)=0 Hence, the possible values of x are −‌
5
2
, and 3 , respectively. Therefore, the sum of the possible values is (3−‌