0=(n-12)(2n+25)

Solution for 0=(n-12)(2n+25) equation:

0=(n-12)(2n+25)

We move all terms to the left:

0-((n-12)(2n+25))=0

We add all the numbers together, and all the variables

-((n-12)(2n+25))=0

We multiply parentheses ..

-((+2n^2+25n-24n-300))=0


We calculate terms in parentheses: -((+2n^2+25n-24n-300)), so:

(+2n^2+25n-24n-300)

We get rid of parentheses

2n^2+25n-24n-300

We add all the numbers together, and all the variables

2n^2+n-300

Back to the equation:

-(2n^2+n-300)


We get rid of parentheses

-2n^2-n+300=0

We add all the numbers together, and all the variables

-2n^2-1n+300=0

a = -2; b = -1; c = +300;
Δ = b2-4ac
Δ = -12-4·(-2)·300
Δ = 2401
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2401}=49$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-49}{2*-2}=\frac{-48}{-4}
=+12
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+49}{2*-2}=\frac{50}{-4}
=-12+1/2
$