2n^2+10n-2=406

Solution for 2n^2+10n-2=406 equation:

2n^2+10n-2=406

We move all terms to the left:

2n^2+10n-2-(406)=0

We add all the numbers together, and all the variables

2n^2+10n-408=0

a = 2; b = 10; c = -408;
Δ = b2-4ac
Δ = 102-4·2·(-408)
Δ = 3364
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{3364}=58$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-58}{2*2}=\frac{-68}{4}
=-17
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+58}{2*2}=\frac{48}{4}
=12
$