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

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
$