text: # (x+3)(x-2)=(x-1)(x+1)
  
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  ## Solution for (x+3)(x-2)=(x-1)(x+1) equation:
  
  (x+3)(x-2)=(x-1)(x+1)
  We move all terms to the left:
  (x+3)(x-2)-((x-1)(x+1))=0
  We use the square of the difference formula
  x^2+(x+3)(x-2)+1=0
  We multiply parentheses ..
  x^2+(+x^2-2x+3x-6)+1=0
  We get rid of parentheses
  x^2+x^2-2x+3x-6+1=0
  We add all the numbers together, and all the variables
  2x^2+x-5=0
  a = 2; b = 1; c = -5;Δ = b2-4acΔ = 12-4·2·(-5)Δ = 41The delta value is higher than zero, so the equation has two solutionsWe use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-\sqrt{41}}{2*2}=\frac{-1-\sqrt{41}}{4}$
  $x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{41}}{2*2}=\frac{-1+\sqrt{41}}{4}$
  
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