Question 29903
-x^2+3x-2=0
Multiplying by (-1)
x^2-3x+2 = 0 ----(1)
x^2-2x-x+2 =0 [splitting the middle term and expressing it as the sum of two terms so that their product is the product of the square term and the constant term. Here (-3x) = (-2x)+(-x) and (-2x)X(-x) = 2x^2 = (x^2)X(2)]
(x^2-2x)-x+2 =0  (by additive associativity)
x(x-2)-1(x-2)=0 (why (-1) pulled out? so that to get(x-2)! )
xp-p= 0 where p= (x-2)
p(x-1) = 0 (taking p out)
(x-2)(x-1) = 0 (putting t back)
(x-2) = 0 gives x = 2
(x-1) = 0 gives x = 1
Answer: x = 1 and x = 2
Verification: x = 1 in (1)
LHS = 1-3+2 = 0 = RHS
x = 2 in (1)
LHS = 4-6+2 = 0=RHS