x^2+22x+109=0

Solution for x^2+22x+109=0 equation:

x^2+22x+109=0

a = 1; b = 22; c = +109;
Δ = b2-4ac
Δ = 222-4·1·109
Δ = 48
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{48}=\sqrt{16*3}=\sqrt{16}*\sqrt{3}=4\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-4\sqrt{3}}{2*1}=\frac{-22-4\sqrt{3}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+4\sqrt{3}}{2*1}=\frac{-22+4\sqrt{3}}{2}
$