Respuesta :

creeds25
x^2-12x+59=0
Using quadratic formula.
[tex]x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a} \\ [/tex]
Here
a=1 , b=-12 and c=59
Put values in formula.
[tex]x=\frac{-\left(-12\right)\pm \sqrt{\left(-12\right)^2-4\cdot \:1\cdot \:59}}{2\cdot \:1} \\ [/tex]
Simplify
[tex]x= \frac{12\pm \sqrt{144-236} }{2} [/tex]
[tex]x=6\pm \sqrt{23}i [/tex]

[tex]Answer: \ 6 \pm \sqrt{23}i [/tex]
carlosego
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)

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