n2+12n=429

Solution for n2+12n=429 equation:

n2+12n=429

We move all terms to the left:

n2+12n-(429)=0

We add all the numbers together, and all the variables

n^2+12n-429=0

a = 1; b = 12; c = -429;
Δ = b2-4ac
Δ = 122-4·1·(-429)
Δ = 1860
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:

$\sqrt{\Delta}=\sqrt{1860}=\sqrt{4*465}=\sqrt{4}*\sqrt{465}=2\sqrt{465}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-2\sqrt{465}}{2*1}=\frac{-12-2\sqrt{465}}{2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+2\sqrt{465}}{2*1}=\frac{-12+2\sqrt{465}}{2}
$