Решите неравенство: 1). 7с(c+2)-c(7c-1)<3. 2). 16x во второй степени - 40x+25>0 Помогите...

$7c(c+2)-c(7c-1)\ \textless \ 3\\7c^2+14c-7c^2+c\ \textless \ 3\\15c\ \textless \ 3\to c\ \textless \ 0,2$
c∈(–∞; $\frac{1}{5}$ )

$16x^2-40x+25\ \textgreater \ 0\\(4x-5)^2\ \textgreater \ 0\to4x-5\ \textgreater \ 0\\4x\ \textgreater \ 5\to x\ \textgreater \ 1,25$
x∈( $1\frac{1}{4}$ ; +∞)