r^2+22r-121=0

Solution for r^2+22r-121=0 equation:

r^2+22r-121=0

a = 1; b = 22; c = -121;
Δ = b2-4ac
Δ = 222-4·1·(-121)
Δ = 968
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:

$\sqrt{\Delta}=\sqrt{968}=\sqrt{484*2}=\sqrt{484}*\sqrt{2}=22\sqrt{2}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-22\sqrt{2}}{2*1}=\frac{-22-22\sqrt{2}}{2}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+22\sqrt{2}}{2*1}=\frac{-22+22\sqrt{2}}{2}
$