s^2-16s-96=0

Solution for s^2-16s-96=0 equation:

s^2-16s-96=0

a = 1; b = -16; c = -96;
Δ = b2-4ac
Δ = -162-4·1·(-96)
Δ = 640
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:

$\sqrt{\Delta}=\sqrt{640}=\sqrt{64*10}=\sqrt{64}*\sqrt{10}=8\sqrt{10}$
$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-8\sqrt{10}}{2*1}=\frac{16-8\sqrt{10}}{2}
$
$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+8\sqrt{10}}{2*1}=\frac{16+8\sqrt{10}}{2}
$