0.1n^2=1

Solution for 0.1n^2=1 equation:

0.1n^2=1

We move all terms to the left:

0.1n^2-(1)=0

a = 0.1; b = 0; c = -1;
Δ = b2-4ac
Δ = 02-4·0.1·(-1)
Δ = 0.4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-\sqrt{0.4}}{2*0.1}=\frac{0-\sqrt{0.4}}{0.2}
=-\frac{\sqrt{}}{0.2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+\sqrt{0.4}}{2*0.1}=\frac{0+\sqrt{0.4}}{0.2}
=\frac{\sqrt{}}{0.2}
$