Answer
{$2 - i\sqrt 5,2 + i\sqrt 5$}
Work Step by Step
Step 1: Comparing $t^{2}-4t+9=0$ to the standard form of a quadratic equation, $at^{2}+bt+c=0$, we find:
$a=1$, $b=-4$ and $c=9$
Step 2: The quadratic formula is:
$x=\frac{-b \pm \sqrt {b^{2}-4ac}}{2a}$
Step 3: Substituting the values of a, b and c in the formula:
$x=\frac{-(-4) \pm \sqrt {(-4)^{2}-4(1)(9)}}{2(1)}$
Step 4: $x=\frac{4 \pm \sqrt {16-36}}{2}$
Step 5: $x=\frac{4 \pm \sqrt {-20}}{2}$
Step 6: $x=\frac{4 \pm \sqrt {-1\times20}}{2}$
Step 7: $x=\frac{4 \pm (\sqrt {-1}\times\sqrt {4\times5})}{2}$
Step 8: $x=\frac{4 \pm (i\times 2\sqrt 5)}{2}$
Step 9: $x=\frac{4 \pm i2\sqrt 5}{2}$
Step 10: $x=\frac{2(2 \pm i\sqrt 5)}{2}$
Step 11: $x=2 \pm i\sqrt 5$
Step 12: $x=2 - i\sqrt 5$ or $x=2 + i\sqrt 5$
Step 13: Therefore, the solution set is {$2 - i\sqrt 5,2 + i\sqrt 5$}.
