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 = -8t^2 - 9t + 8 = 0(t - 8)(t - 1) = 0
Therefore, t = 8 or t = 1 when a = -8
For a2 = -14:
t^2 - 9t = -14t^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.
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.