0=t^2-12.2t+18

Solution for 0=t^2-12.2t+18 equation:

0=t^2-12.2t+18

We move all terms to the left:

0-(t^2-12.2t+18)=0

We add all the numbers together, and all the variables

-(t^2-12.2t+18)=0

We get rid of parentheses

-t^2+12.2t-18=0

We add all the numbers together, and all the variables

-1t^2+12.2t-18=0

a = -1; b = 12.2; c = -18;
Δ = b2-4ac
Δ = 12.22-4·(-1)·(-18)
Δ = 76.84
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12.2)-\sqrt{76.84}}{2*-1}=\frac{-12.2-\sqrt{76.84}}{-2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12.2)+\sqrt{76.84}}{2*-1}=\frac{-12.2+\sqrt{76.84}}{-2}
$