15=18*t-0,5t^2

Solution for 15=18*t-0,5t^2 equation:

15=18t-0.5t^2

We move all terms to the left:

15-(18t-0.5t^2)=0

We get rid of parentheses

0.5t^2-18t+15=0

a = 0.5; b = -18; c = +15;
Δ = b2-4ac
Δ = -182-4·0.5·15
Δ = 294
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:

$\sqrt{\Delta}=\sqrt{294}=\sqrt{49*6}=\sqrt{49}*\sqrt{6}=7\sqrt{6}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-7\sqrt{6}}{2*0.5}=\frac{18-7\sqrt{6}}{1}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+7\sqrt{6}}{2*0.5}=\frac{18+7\sqrt{6}}{1}
$