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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

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


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

4(+X)

We multiply parentheses

4X

Back to the equation:

-(4X)


We get rid of parentheses

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

We add all the numbers together, and all the variables

-1X^2+41X-200=0

a = -1; b = 41; c = -200;
Δ = b2-4ac
Δ = 412-4·(-1)·(-200)
Δ = 881
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(41)-\sqrt{881}}{2*-1}=\frac{-41-\sqrt{881}}{-2}
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(41)+\sqrt{881}}{2*-1}=\frac{-41+\sqrt{881}}{-2}
$