r^2-2.5r+1=0

Solution for r^2-2.5r+1=0 equation:

r^2-2.5r+1=0

a = 1; b = -2.5; c = +1;
Δ = b2-4ac
Δ = -2.52-4·1·1
Δ = 2.25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2.5)-\sqrt{2.25}}{2*1}=\frac{2.5-\sqrt{2.25}}{2}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2.5)+\sqrt{2.25}}{2*1}=\frac{2.5+\sqrt{2.25}}{2}
$