0=1.8t-5(t^2)

Solution for 0=1.8t-5(t^2) equation:

0=1.8t-5(t^2)

We move all terms to the left:

0-(1.8t-5(t^2))=0

We add all the numbers together, and all the variables

-(1.8t-5t^2)=0

We get rid of parentheses

5t^2-1.8t=0

a = 5; b = -1.8; c = 0;
Δ = b2-4ac
Δ = -1.82-4·5·0
Δ = 3.24
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{-(-1.8)-\sqrt{3.24}}{2*5}=\frac{1.8-\sqrt{3.24}}{10}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1.8)+\sqrt{3.24}}{2*5}=\frac{1.8+\sqrt{3.24}}{10}
$