3n=(n-4)(n-5)

Solution for 3n=(n-4)(n-5) equation:

3n=(n-4)(n-5)

We move all terms to the left:

3n-((n-4)(n-5))=0

We multiply parentheses ..

-((+n^2-5n-4n+20))+3n=0


We calculate terms in parentheses: -((+n^2-5n-4n+20)), so:

(+n^2-5n-4n+20)

We get rid of parentheses

n^2-5n-4n+20

We add all the numbers together, and all the variables

n^2-9n+20

Back to the equation:

-(n^2-9n+20)


We add all the numbers together, and all the variables

3n-(n^2-9n+20)=0

We get rid of parentheses

-n^2+3n+9n-20=0

We add all the numbers together, and all the variables

-1n^2+12n-20=0

a = -1; b = 12; c = -20;
Δ = b2-4ac
Δ = 122-4·(-1)·(-20)
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-8}{2*-1}=\frac{-20}{-2}
=+10
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+8}{2*-1}=\frac{-4}{-2}
=+2
$