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

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

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

We move all terms to the left:

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

We multiply parentheses ..

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


We calculate terms in parentheses: -(4(X-40)), so:

4(X-40)

We multiply parentheses

4X-160

Back to the equation:

-(4X-160)


We get rid of parentheses

X^2-40X-5X-4X+200+160=0

We add all the numbers together, and all the variables

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