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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

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


We calculate terms in parentheses: -(4(x-5)), so:

4(x-5)

We multiply parentheses

4x-20

Back to the equation:

-(4x-20)


We get rid of parentheses

-1x^2+40x+5x-4x-200+20=0

We add all the numbers together, and all the variables

-1x^2+41x-180=0

a = -1; b = 41; c = -180;
Δ = b2-4ac
Δ = 412-4·(-1)·(-180)
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{-(41)-31}{2*-1}=\frac{-72}{-2}
=+36
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(41)+31}{2*-1}=\frac{-10}{-2}
=+5
$