0=2t^2-9t-18

Solution for 0=2t^2-9t-18 equation:

0=2t^2-9t-18

We move all terms to the left:

0-(2t^2-9t-18)=0

We add all the numbers together, and all the variables

-(2t^2-9t-18)=0

We get rid of parentheses

-2t^2+9t+18=0

a = -2; b = 9; c = +18;
Δ = b2-4ac
Δ = 92-4·(-2)·18
Δ = 225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{225}=15$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-15}{2*-2}=\frac{-24}{-4}
=+6
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+15}{2*-2}=\frac{6}{-4}
=-1+1/2
$