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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

(x-4)(x-3)-(-1x+7)=0

We get rid of parentheses

(x-4)(x-3)+1x-7=0

We multiply parentheses ..

(+x^2-3x-4x+12)+1x-7=0

We add all the numbers together, and all the variables

(+x^2-3x-4x+12)+x-7=0

We get rid of parentheses

x^2-3x-4x+x+12-7=0

We add all the numbers together, and all the variables

x^2-6x+5=0

a = 1; b = -6; c = +5;
Δ = b2-4ac
Δ = -62-4·1·5
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-4}{2*1}=\frac{2}{2}
=1
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+4}{2*1}=\frac{10}{2}
=5
$