0=18t-t^2+6

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

0=18t-t^2+6

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

t^2-18t-6=0

a = 1; b = -18; c = -6;
Δ = b2-4ac
Δ = -182-4·1·(-6)
Δ = 348
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:

$\sqrt{\Delta}=\sqrt{348}=\sqrt{4*87}=\sqrt{4}*\sqrt{87}=2\sqrt{87}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-2\sqrt{87}}{2*1}=\frac{18-2\sqrt{87}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+2\sqrt{87}}{2*1}=\frac{18+2\sqrt{87}}{2}
$