How do I make this app factor a polynomial with quadratic formula or some other trick? - polynomials

Question

How do I make this app factor a polynomial with quadratic formula or some other trick?.
The polynomial 3x^2 +10x -8 can be factored to (x +4)(3x -2) with the use of
grouping or the X trick. I would like my code to do this
but I have no clue how to accomplish it. My code factors the polynomial to (x -0.6666666666666666)(x +4.0)
which is not what I want from it.
This code snippet
require "option_parser"
puts "Please enter your variable values assuming the form ax**2 + bx + c ."
puts "a: "
a = gets
exit if a.nil?
a = a.to_i
puts "b: "
b = gets
exit if b.nil?
b = b.to_i
puts "c: "
c = gets
exit if c.nil?
c = c.to_i
the_gcd = a.gcd(b).gcd(a.gcd(c))
if the_gcd != 1
x1 = (-1*b + (b**2 - 4*a*c) ** 0.5)/(2 * a)
x2 = (-1*b - (b**2 - 4*a*c) ** 0.5)/(2 * a)
if x1.to_i - x1 != 0 || x2.to_i - x2 != 0
puts "The root is not a whole number. Consider grouping for factoring the polynomial."
if x1 < 0 && x2 < 0
puts "#{the_gcd}(x +#{-x1})(x +#{-x2})"
elsif x1 < 0 && x2 > 0
puts "#{the_gcd}(x +#{-x1})(x #{-x2})"
elsif x1 > 0 && x2 < 0
puts "#{the_gcd}(x #{-x1})(x +#{-x2})"
elsif x1 > 0 && x2 > 0
puts "#{the_gcd}(x #{-x1})(x #{-x2})"
end
exit
else
if x1 < 0 && x2 < 0
puts "#{the_gcd}(x +#{-x1.to_i})(x +#{-x2.to_i})"
exit
elsif x1 < 0 && x2 > 0
puts "#{the_gcd}(x +#{-x1.to_i})(x #{-x2.to_i})"
exit
elsif x1 > 0 && x2 < 0
puts "#{the_gcd}(x #{-x1.to_i})(x +#{-x2.to_i})"
exit
elsif x1 > 0 && x2 > 0
puts "#{the_gcd}(x #{-x1.to_i})(x #{-x2.to_i})"
exit
end
end
if (b**2 - 4*a*c) < 0
puts "No real solution. Imaginary numbers involved in the solution."
exit
end
end
#The part below does not utilize GCD
x1 = (-1*b + (b**2 - 4*a*c) ** 0.5)/(2 * a)
x2 = (-1*b - (b**2 - 4*a*c) ** 0.5)/(2 * a)
if x1.to_i - x1 != 0 || x2.to_i - x2 != 0
puts "The root is not a whole number. Consider grouping for factoring the polynomial."
if x1 < 0 && x2 < 0
puts "(x +#{-x1})(x +#{-x2})"
elsif x1 < 0 && x2 > 0
puts "(x +#{-x1})(x #{-x2})"
elsif x1 > 0 && x2 < 0
puts "(x #{-x1})(x +#{-x2})"
elsif x1 > 0 && x2 > 0
puts "(x #{-x1})(x #{-x2})"
end
exit
else
if x1 < 0 && x2 < 0
puts "(x +#{-x1.to_i})(x +#{-x2.to_i})"
elsif x1 < 0 && x2 > 0
puts "(x +#{-x1.to_i})(x #{-x2.to_i})"
elsif x1 > 0 && x2 < 0
puts "(x #{-x1.to_i})(x +#{-x2.to_i})"
elsif x1 > 0 && x2 > 0
puts "(x #{-x1.to_i})(x #{-x2.to_i})"
end
end
if (b**2 - 4*a*c) < 0
puts "No real solution. Imaginary numbers involved in the solution."
exit
end
is part of a larger app that deals with other methods of factoring quadratic polynomials as well.
Let these other methods of factoring, such as factoring perfect square trinomials or difference of perfect squares, be
outside of the scope of this question.
I came across a similar problem in this Python code, that apparently was solved with the use of fractions.
Might one utilize fractions in Crystal in order to solve my problem?