5t^2-25t+0.2=0

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

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}
$