find the valuef k

find the values of k if the expression 4x² + 2(k + 3)x + 9k² is positive for all real values of x.

if ax²+bx+c is always positive, Δ＜0 and a＞0
in the equation 4x² + 2(k + 3)x + 9k²
Δ=[2(k+3)]²-4(4)(9k²)＜０
[2(k+3)]²＜144k²
2(k+3)＜12k
2k+6＜12k
6＜10k
k＞3/5

for a quadratic function to have 1 or 2 solution then
b²-4ac＞0
∴ [2(k + 3)]² - 4×4×9k²＞0
4k²+24k+36-144k²＞0
-140k²+24k+36＞0

use,
-b±√(b²-4ac)÷(2a) to factorize -140k²+24k+36 give
(k+3/7)(k-3/5)＞0
k＞-3/7　∩　k＞3/5　　　　Note:　∩ mean *and* or *intercept*
∴k＞3/5

AMN46KR9497@57M:
u missed the important point where -140k²+24k+36 give two values not one. The correct way to solve a quadratic function must factorize it to give two sovle