r^2+22r+105=0

Solution for r^2+22r+105=0 equation:

r^2+22r+105=0

a = 1; b = 22; c = +105;
Δ = b2-4ac
Δ = 222-4·1·105
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-8}{2*1}=\frac{-30}{2}
=-15
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+8}{2*1}=\frac{-14}{2}
=-7
$