中四數學問題

ax^2 + bx + c = 0
有不相同的實根時,, b^2 - 4ac > 0
有相同實根時,, b^2 - 4ac = 0
沒有實根時,, b^2 - 4ac < 0

所以

#1 2x^2 + 8x + (3-k) = 0

b^2 - 4ac > 0
8^2 - 4(2)(3-k) > 0
64 - 24 + 8k > 0
8k > - 40
k > - 5

#2 2x^2 + kx + (k+6) =0

b^2 - 4ac = 0
k^2 - 4(2)(k+6) = 0
k^2 - 8k - 48 = 0
( k - 12 ) ( k + 4 ) = 0
所以,, k = 12 或 k = - 4

First Question
(8)^2-4(2)(3-k) >0
64-24+8k>0
8k>-40
k>-5

Second Question
(k)^2-4(2)(k+6)=0
k^2-8k-48=0
k=-12 or 4