r^2+8r+11=0

Solution for r^2+8r+11=0 equation:

r^2+8r+11=0

a = 1; b = 8; c = +11;
Δ = b2-4ac
Δ = 82-4·1·11
Δ = 20
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{20}=\sqrt{4*5}=\sqrt{4}*\sqrt{5}=2\sqrt{5}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-2\sqrt{5}}{2*1}=\frac{-8-2\sqrt{5}}{2}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+2\sqrt{5}}{2*1}=\frac{-8+2\sqrt{5}}{2}
$