Question

Найдите корни уравнения :25-(13+ t)^2=0

Answer 1

25-(13+ t)² = 0;

25-(169+26t+t²) = 0;
25-169-26t-t² = 0;
-t²-26t-144 = 0;
D = b²-4ac = 676-4*(-1)*(-144) = 100.

t1 = -b+√D/2a = 26+10/(-2) = 36/(-2) = -18;
t2 = -b-√D/2a = 26-10/(-2) = 16/(-2) = -8.

Ответ: t1 = -18; t2 = -8.