(40-x)(x-5)=4(40-x)

Solution for (40-x)(x-5)=4(40-x) equation:

(40-x)(x-5)=4(40-x)

We move all terms to the left:

(40-x)(x-5)-(4(40-x))=0

We add all the numbers together, and all the variables

(-1x+40)(x-5)-(4(-1x+40))=0

We multiply parentheses ..

(-1x^2+5x+40x-200)-(4(-1x+40))=0


We calculate terms in parentheses: -(4(-1x+40)), so:

4(-1x+40)

We multiply parentheses

-4x+160

Back to the equation:

-(-4x+160)


We get rid of parentheses

-1x^2+5x+40x+4x-200-160=0

We add all the numbers together, and all the variables

-1x^2+49x-360=0

a = -1; b = 49; c = -360;
Δ = b2-4ac
Δ = 492-4·(-1)·(-360)
Δ = 961
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{961}=31$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(49)-31}{2*-1}=\frac{-80}{-2}
=+40
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(49)+31}{2*-1}=\frac{-18}{-2}
=+9
$