s^2+2.5s+0.5=0

Solution for s^2+2.5s+0.5=0 equation:

s^2+2.5s+0.5=0

a = 1; b = 2.5; c = +0.5;
Δ = b2-4ac
Δ = 2.52-4·1·0.5
Δ = 4.25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2.5)-\sqrt{4.25}}{2*1}=\frac{-2.5-\sqrt{4.25}}{2}
$
$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2.5)+\sqrt{4.25}}{2*1}=\frac{-2.5+\sqrt{4.25}}{2}
$