0=-t^2+11t

Solution for 0=-t^2+11t equation:

0=-t^2+11t

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

t^2-11t=0

a = 1; b = -11; c = 0;
Δ = b2-4ac
Δ = -112-4·1·0
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-11}{2*1}=\frac{0}{2}
=0
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+11}{2*1}=\frac{22}{2}
=11
$