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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(-3x^2+7x+12x-28)-(x(x-4))=0


We calculate terms in parentheses: -(x(x-4)), so:

x(x-4)

We multiply parentheses

x^2-4x

Back to the equation:

-(x^2-4x)


We get rid of parentheses

-3x^2-x^2+7x+12x+4x-28=0

We add all the numbers together, and all the variables

-4x^2+23x-28=0

a = -4; b = 23; c = -28;
Δ = b2-4ac
Δ = 232-4·(-4)·(-28)
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(23)-9}{2*-4}=\frac{-32}{-8}
=+4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+9}{2*-4}=\frac{-14}{-8}
=1+3/4
$