(x-1)(x-2)+(x-2)(x-3)+(x-3)(x-4)=0

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Solution for (x-1)(x-2)+(x-2)(x-3)+(x-3)(x-4)=0 equation:



(x-1)(x-2)+(x-2)(x-3)+(x-3)(x-4)=0
We multiply parentheses ..
(+x^2-2x-1x+2)+(x-2)(x-3)+(x-3)(x-4)=0
We get rid of parentheses
x^2-2x-1x+(x-2)(x-3)+(x-3)(x-4)+2=0
We multiply parentheses ..
x^2+(+x^2-3x-2x+6)-2x-1x+(x-3)(x-4)+2=0
We add all the numbers together, and all the variables
x^2+(+x^2-3x-2x+6)-3x+(x-3)(x-4)+2=0
We get rid of parentheses
x^2+x^2-3x-2x-3x+(x-3)(x-4)+6+2=0
We multiply parentheses ..
x^2+x^2+(+x^2-4x-3x+12)-3x-2x-3x+6+2=0
We add all the numbers together, and all the variables
2x^2+(+x^2-4x-3x+12)-8x+8=0
We get rid of parentheses
2x^2+x^2-4x-3x-8x+12+8=0
We add all the numbers together, and all the variables
3x^2-15x+20=0
a = 3; b = -15; c = +20;
Δ = b2-4ac
Δ = -152-4·3·20
Δ = -15
Delta is less than zero, so there is no solution for the equation

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