0=5t^2-25t+0.2

Solution for 0=5t^2-25t+0.2 equation:

0=5t^2-25t+0.2

We move all terms to the left:

0-(5t^2-25t+0.2)=0

We add all the numbers together, and all the variables

-(5t^2-25t+0.2)=0

We get rid of parentheses

-5t^2+25t-0.2=0

a = -5; b = 25; c = -0.2;
Δ = b2-4ac
Δ = 252-4·(-5)·(-0.2)
Δ = 621
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{621}=\sqrt{9*69}=\sqrt{9}*\sqrt{69}=3\sqrt{69}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(25)-3\sqrt{69}}{2*-5}=\frac{-25-3\sqrt{69}}{-10}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+3\sqrt{69}}{2*-5}=\frac{-25+3\sqrt{69}}{-10}
$