a^2+22a-105=0

Solution for a^2+22a-105=0 equation:

a^2+22a-105=0

a = 1; b = 22; c = -105;
Δ = b2-4ac
Δ = 222-4·1·(-105)
Δ = 904
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{904}=\sqrt{4*226}=\sqrt{4}*\sqrt{226}=2\sqrt{226}$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-2\sqrt{226}}{2*1}=\frac{-22-2\sqrt{226}}{2}
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+2\sqrt{226}}{2*1}=\frac{-22+2\sqrt{226}}{2}
$