(x-4)(x-3)=x-7x+12

Solution for (x-4)(x-3)=x-7x+12 equation:

(x-4)(x-3)=x-7x+12

We move all terms to the left:

(x-4)(x-3)-(x-7x+12)=0

We add all the numbers together, and all the variables

(x-4)(x-3)-(-6x+12)=0

We get rid of parentheses

(x-4)(x-3)+6x-12=0

We multiply parentheses ..

(+x^2-3x-4x+12)+6x-12=0

We get rid of parentheses

x^2-3x-4x+6x+12-12=0

We add all the numbers together, and all the variables

x^2-1x=0

a = 1; b = -1; c = 0;
Δ = b2-4ac
Δ = -12-4·1·0
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-1}{2*1}=\frac{0}{2}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+1}{2*1}=\frac{2}{2}
=1
$