t^2-10t-96=0

Solution for t^2-10t-96=0 equation:

t^2-10t-96=0

a = 1; b = -10; c = -96;
Δ = b2-4ac
Δ = -102-4·1·(-96)
Δ = 484
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{484}=22$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-22}{2*1}=\frac{-12}{2}
=-6
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+22}{2*1}=\frac{32}{2}
=16
$