((n-1)(n-2)=30)

Solution for ((n-1)(n-2)=30) equation:

((n-1)(n-2)=30)

We move all terms to the left:

((n-1)(n-2)-(30))=0

We multiply parentheses ..

((+n^2-2n-1n+2)-30)=0


We calculate terms in parentheses: +((+n^2-2n-1n+2)-30), so:

(+n^2-2n-1n+2)-30

We get rid of parentheses

n^2-2n-1n+2-30

We add all the numbers together, and all the variables

n^2-3n-28

Back to the equation:

+(n^2-3n-28)


We get rid of parentheses

n^2-3n-28=0

a = 1; b = -3; c = -28;
Δ = b2-4ac
Δ = -32-4·1·(-28)
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-11}{2*1}=\frac{-8}{2}
=-4
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+11}{2*1}=\frac{14}{2}
=7
$