0=-0.2t^2+5t-17

Solution for 0=-0.2t^2+5t-17 equation:

0=-0.2t^2+5t-17

We move all terms to the left:

0-(-0.2t^2+5t-17)=0

We add all the numbers together, and all the variables

-(-0.2t^2+5t-17)=0

We get rid of parentheses

0.2t^2-5t+17=0

a = 0.2; b = -5; c = +17;
Δ = b2-4ac
Δ = -52-4·0.2·17
Δ = 11.4
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{-(-5)-\sqrt{11.4}}{2*0.2}=\frac{5-\sqrt{11.4}}{0.4}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+\sqrt{11.4}}{2*0.2}=\frac{5+\sqrt{11.4}}{0.4}
$