Question

(t^2-9t)^2+22(t^2-9t)+112=0

Answer 1

t2-9t=aa2+22a+112=0
delta=121-112=9a=-11+3=-8  a2=-11-3=-14t2-9t=-8t2-9t+8=0
D=81-32=49   t1=9+7/2=8    t2=9-7/2=1    t2-9t=-14t2-9t+14=0D=81-56=25  t1=9+5/2=7  t2=9-5/2=2   КАК ТО ТАК

Answer 2

$(t^2-9t)^2+22(t^2-9t)+112=0$
пусть $(t^2-9t) = x$, тогда
$x^2+22x+112=0$
$x_{1,2}= frac{-bpmsqrt{D}}{2a} = frac{-22pmsqrt{36}}{2} = frac{-22pm6}{2}$
$x_1 = -8,; x_2 = -14$
тогда
$t^2-9t = -8$ решаем уравнение $t^2-9t + 8 =0$
$t_{1,2}= frac{-bpmsqrt{D}}{2a} = frac{9pmsqrt{49}}{2} = frac{9pm7}{2}$
$t_1 = 8,; t_2 = 1$
и
$t^2-9t = -4$ решаем уравнение $t^2-9t + 4 =0$
$t_{3,4}= frac{-bpmsqrt{D}}{2a} = frac{9pmsqrt{65}}{2} = frac{9pmsqrt{65}}{2}$
$t_3 = frac{9+sqrt{65}}{2},; t_4 = frac{9-sqrt{65}}{2}$

Ответ:
$t_1 = 8,; t_2 = 1$
$t_3 = frac{9+sqrt{65}}{2},; t_4 = frac{9-sqrt{65}}{2}$