0.5n^2+1.5n+1=0

Solution for 0.5n^2+1.5n+1=0 equation:

0.5n^2+1.5n+1=0

a = 0.5; b = 1.5; c = +1;
Δ = b2-4ac
Δ = 1.52-4·0.5·1
Δ = 0.25
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{-(1.5)-\sqrt{0.25}}{2*0.5}=\frac{-1.5-\sqrt{0.25}}{1}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1.5)+\sqrt{0.25}}{2*0.5}=\frac{-1.5+\sqrt{0.25}}{1}
$