t ≥ 0
36t2+35t–1 ≤ 0
D=352–4·36·(–1)=1225+144=1369=372
t1= (–35–37)/72= –1 или t2=(–35+37)/72=1/36
Решение неравенства
36t2+35t–1 ≤ 0
–1 ≤ t ≤ 1/36
с учетом t ≥ 0
0 ≤t ≤ 1/36
Обратный переход
0 ≤x2 ≤ 1/36
x2 ≥ 0 при любом х
x2 ≤ 1/36 ⇔ |x| ≤ 1/6 ⇔ –1/6 ≤ x ≤ 1/6