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

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

(X-5)(40-X)=(X-5)4

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

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


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

(X-5)4

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
$