r2-5r+2=0

Solution for r2-5r+2=0 equation:

r2-5r+2=0

We add all the numbers together, and all the variables

r^2-5r+2=0

a = 1; b = -5; c = +2;
Δ = b2-4ac
Δ = -52-4·1·2
Δ = 17
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{-(-5)-\sqrt{17}}{2*1}=\frac{5-\sqrt{17}}{2}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+\sqrt{17}}{2*1}=\frac{5+\sqrt{17}}{2}
$