0=(2n-63)(n-30)

Solution for 0=(2n-63)(n-30) equation:

0=(2n-63)(n-30)

We move all terms to the left:

0-((2n-63)(n-30))=0

We add all the numbers together, and all the variables

-((2n-63)(n-30))=0

We multiply parentheses ..

-((+2n^2-60n-63n+1890))=0


We calculate terms in parentheses: -((+2n^2-60n-63n+1890)), so:

(+2n^2-60n-63n+1890)

We get rid of parentheses

2n^2-60n-63n+1890

We add all the numbers together, and all the variables

2n^2-123n+1890

Back to the equation:

-(2n^2-123n+1890)


We get rid of parentheses

-2n^2+123n-1890=0

a = -2; b = 123; c = -1890;
Δ = b2-4ac
Δ = 1232-4·(-2)·(-1890)
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{-(123)-3}{2*-2}=\frac{-126}{-4}
=31+1/2
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(123)+3}{2*-2}=\frac{-120}{-4}
=+30
$