# (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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