l^2-22l-133=0

Solution for l^2-22l-133=0 equation:

l^2-22l-133=0

a = 1; b = -22; c = -133;
Δ = b2-4ac
Δ = -222-4·1·(-133)
Δ = 1016
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$l_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$l_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1016}=\sqrt{4*254}=\sqrt{4}*\sqrt{254}=2\sqrt{254}$
$l_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-22)-2\sqrt{254}}{2*1}=\frac{22-2\sqrt{254}}{2}
$
$l_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-22)+2\sqrt{254}}{2*1}=\frac{22+2\sqrt{254}}{2}
$