1n^2+12n+3=0

Solution for 1n^2+12n+3=0 equation:

1n^2+12n+3=0

We add all the numbers together, and all the variables

n^2+12n+3=0

a = 1; b = 12; c = +3;
Δ = b2-4ac
Δ = 122-4·1·3
Δ = 132
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:

$\sqrt{\Delta}=\sqrt{132}=\sqrt{4*33}=\sqrt{4}*\sqrt{33}=2\sqrt{33}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-2\sqrt{33}}{2*1}=\frac{-12-2\sqrt{33}}{2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+2\sqrt{33}}{2*1}=\frac{-12+2\sqrt{33}}{2}
$