t^2+100t-9600=0

Solution for t^2+100t-9600=0 equation:

t^2+100t-9600=0

a = 1; b = 100; c = -9600;
Δ = b2-4ac
Δ = 1002-4·1·(-9600)
Δ = 48400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{48400}=220$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(100)-220}{2*1}=\frac{-320}{2}
=-160
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(100)+220}{2*1}=\frac{120}{2}
=60
$