0=t^2-12t-108

Solution for 0=t^2-12t-108 equation:

0=t^2-12t-108

We move all terms to the left:

0-(t^2-12t-108)=0

We add all the numbers together, and all the variables

-(t^2-12t-108)=0

We get rid of parentheses

-t^2+12t+108=0

We add all the numbers together, and all the variables

-1t^2+12t+108=0

a = -1; b = 12; c = +108;
Δ = b2-4ac
Δ = 122-4·(-1)·108
Δ = 576
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{576}=24$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-24}{2*-1}=\frac{-36}{-2}
=+18
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+24}{2*-1}=\frac{12}{-2}
=-6
$