0=t^2-6t-18

Solution for 0=t^2-6t-18 equation:

0=t^2-6t-18

We move all terms to the left:

0-(t^2-6t-18)=0

We add all the numbers together, and all the variables

-(t^2-6t-18)=0

We get rid of parentheses

-t^2+6t+18=0

We add all the numbers together, and all the variables

-1t^2+6t+18=0

a = -1; b = 6; c = +18;
Δ = b2-4ac
Δ = 62-4·(-1)·18
Δ = 108
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{108}=\sqrt{36*3}=\sqrt{36}*\sqrt{3}=6\sqrt{3}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-6\sqrt{3}}{2*-1}=\frac{-6-6\sqrt{3}}{-2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+6\sqrt{3}}{2*-1}=\frac{-6+6\sqrt{3}}{-2}
$