Question

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Answer 1

in this case it is to find the roots of the polynomial.
We have then:
x ^ 2-12x + 59 = 0
Applying resolver we have
x = (- b +/- root (b ^ 2 - 4ac)) / (2a)
Substituting values:
x = (- (- 12) +/- root ((- 12) ^ 2 - 4 (1) (59))) / (2 (1))
x = (- (- 12) +/- root ((144 - 236)) / (2 (1))
x = (12 +/- root (-92))) / (2)
x = (6 +/- i * root (23)))
Answer:
x = (6 +/- i * root (23)))
(option 3)

Answer 2

x^2-12x+59=0
Using quadratic formula.

`x_(1,\:2)=(-b\pm √(b^2-4ac))/(2a) \\`
Here
a=1 , b=-12 and c=59
Put values in formula.

`x=(-\left(-12\right)\pm √(\left(-12\right)^2-4\cdot \:1\cdot \:59))/(2\cdot \:1) \\`
Simplify

`x= (12\pm √(144-236) )/(2)`

`x=6\pm √(23)i`


`Answer: \ 6 \pm √(23)i`