0=(n-2)(n-5)

Solution for 0=(n-2)(n-5) equation:

0=(n-2)(n-5)

We move all terms to the left:

0-((n-2)(n-5))=0

We add all the numbers together, and all the variables

-((n-2)(n-5))=0

We multiply parentheses ..

-((+n^2-5n-2n+10))=0


We calculate terms in parentheses: -((+n^2-5n-2n+10)), so:

(+n^2-5n-2n+10)

We get rid of parentheses

n^2-5n-2n+10

We add all the numbers together, and all the variables

n^2-7n+10

Back to the equation:

-(n^2-7n+10)


We get rid of parentheses

-n^2+7n-10=0

We add all the numbers together, and all the variables

-1n^2+7n-10=0

a = -1; b = 7; c = -10;
Δ = b2-4ac
Δ = 72-4·(-1)·(-10)
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-3}{2*-1}=\frac{-10}{-2}
=+5
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+3}{2*-1}=\frac{-4}{-2}
=+2
$