x^2+22x-384=0

Solution for x^2+22x-384=0 equation:

x^2+22x-384=0

a = 1; b = 22; c = -384;
Δ = b2-4ac
Δ = 222-4·1·(-384)
Δ = 2020
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:

$\sqrt{\Delta}=\sqrt{2020}=\sqrt{4*505}=\sqrt{4}*\sqrt{505}=2\sqrt{505}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-2\sqrt{505}}{2*1}=\frac{-22-2\sqrt{505}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+2\sqrt{505}}{2*1}=\frac{-22+2\sqrt{505}}{2}
$