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

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вопрос

To solve the quadratic equation, we first need to let a = (t^2 - 9t).

Substitute a into the equation, we get:

a^2 + 22a + 112 = 0

Now, we have a quadratic equation in the form of ax^2 + bx + c = 0. We can solve it using the quadratic formula:

a = (-b ± sqrt(b^2 - 4ac)) / 2a

Now, plug in the values of a, b, and c:

a = -22 ± sqrt(22^2 - 41112) / 2

a = -22 ± sqrt(484 - 448) / 2

a = -22 ± sqrt(36) / 2

a = (-22 ± 6) / 2

Now, we have two possible solutions for a:

a1 = (-22 + 6) / 2 = -16 / 2 = -8

a2 = (-22 - 6) / 2 = -28 / 2 = -14

Now, we substitute back in for a = (t^2 - 9t) to find the values of t:

For a1 = -8:

t^2 - 9t = -8
t^2 - 9t + 8 = 0
(t - 8)(t - 1) = 0

Therefore, t = 8 or t = 1 when a = -8

For a2 = -14:

t^2 - 9t = -14
t^2 - 9t + 14 = 0
(t - 7)(t - 2) = 0

Therefore, t = 7 or t = 2 when a = -14

Therefore, the solutions to the given quadratic equation are t = 8, t = 1, t = 7, and t = 2.