12n=n^2+19

Solution for 12n=n^2+19 equation:

12n=n^2+19

We move all terms to the left:

12n-(n^2+19)=0

We get rid of parentheses

-n^2+12n-19=0

We add all the numbers together, and all the variables

-1n^2+12n-19=0

a = -1; b = 12; c = -19;
Δ = b2-4ac
Δ = 122-4·(-1)·(-19)
Δ = 68
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{68}=\sqrt{4*17}=\sqrt{4}*\sqrt{17}=2\sqrt{17}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-2\sqrt{17}}{2*-1}=\frac{-12-2\sqrt{17}}{-2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+2\sqrt{17}}{2*-1}=\frac{-12+2\sqrt{17}}{-2}
$