10n2+1=491

Solution for 10n2+1=491 equation:

10n^2+1=491

We move all terms to the left:

10n^2+1-(491)=0

We add all the numbers together, and all the variables

10n^2-490=0

a = 10; b = 0; c = -490;
Δ = b2-4ac
Δ = 02-4·10·(-490)
Δ = 19600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{19600}=140$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-140}{2*10}=\frac{-140}{20}
=-7
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+140}{2*10}=\frac{140}{20}
=7
$