0=2912t-18.2t^2

Solution for 0=2912t-18.2t^2 equation:

0=2912t-18.2t^2

We move all terms to the left:

0-(2912t-18.2t^2)=0

We add all the numbers together, and all the variables

-(2912t-18.2t^2)=0

We get rid of parentheses

18.2t^2-2912t=0

a = 18.2; b = -2912; c = 0;
Δ = b2-4ac
Δ = -29122-4·18.2·0
Δ = 8479744
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{8479744}=2912$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2912)-2912}{2*18.2}=\frac{0}{36.4}
=0
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2912)+2912}{2*18.2}=\frac{5824}{36.4}
=160
$