12n=n^2+35

Solution for 12n=n^2+35 equation:

12n=n^2+35

We move all terms to the left:

12n-(n^2+35)=0

We get rid of parentheses

-n^2+12n-35=0

We add all the numbers together, and all the variables

-1n^2+12n-35=0

a = -1; b = 12; c = -35;
Δ = b2-4ac
Δ = 122-4·(-1)·(-35)
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-2}{2*-1}=\frac{-14}{-2}
=+7
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+2}{2*-1}=\frac{-10}{-2}
=+5
$