r2+5r+1=0

Solution for r2+5r+1=0 equation:

r2+5r+1=0

We add all the numbers together, and all the variables

r^2+5r+1=0

a = 1; b = 5; c = +1;
Δ = b2-4ac
Δ = 52-4·1·1
Δ = 21
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{21}}{2*1}=\frac{-5-\sqrt{21}}{2}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+\sqrt{21}}{2*1}=\frac{-5+\sqrt{21}}{2}
$