s^2+22s-1600=0

Solution for s^2+22s-1600=0 equation:

s^2+22s-1600=0

a = 1; b = 22; c = -1600;
Δ = b2-4ac
Δ = 222-4·1·(-1600)
Δ = 6884
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{6884}=\sqrt{4*1721}=\sqrt{4}*\sqrt{1721}=2\sqrt{1721}$
$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-2\sqrt{1721}}{2*1}=\frac{-22-2\sqrt{1721}}{2}
$
$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+2\sqrt{1721}}{2*1}=\frac{-22+2\sqrt{1721}}{2}
$