11=0t+.5t^2

Solution for 11=0t+.5t^2 equation:

11=0t+.5t^2

We move all terms to the left:

11-(0t+.5t^2)=0

We get rid of parentheses

-.5t^2-0t+11=0

We add all the numbers together, and all the variables

-0.5t^2-1t+11=0

a = -0.5; b = -1; c = +11;
Δ = b2-4ac
Δ = -12-4·(-0.5)·11
Δ = 23
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{-(-1)-\sqrt{23}}{2*-0.5}=\frac{1-\sqrt{23}}{-1}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{23}}{2*-0.5}=\frac{1+\sqrt{23}}{-1}
$