(x-4)(x-3)=(x-3)2-(x-3)

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



(x-4)(x-3)=(x-3)2-(x-3)
We move all terms to the left:
(x-4)(x-3)-((x-3)2-(x-3))=0
We multiply parentheses ..
(+x^2-3x-4x+12)-((x-3)2-(x-3))=0
We calculate terms in parentheses: -((x-3)2-(x-3)), so:
(x-3)2-(x-3)
We multiply parentheses
2x-(x-3)-6
We get rid of parentheses
2x-x+3-6
We add all the numbers together, and all the variables
x-3
Back to the equation:
-(x-3)
We get rid of parentheses
x^2-3x-4x-x+12+3=0
We add all the numbers together, and all the variables
x^2-8x+15=0
a = 1; b = -8; c = +15;
Δ = b2-4ac
Δ = -82-4·1·15
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-2}{2*1}=\frac{6}{2} =3 $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+2}{2*1}=\frac{10}{2} =5 $

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