0.20t^2-25=0

Solution for 0.20t^2-25=0 equation:

0.20t^2-25=0

a = 0.20; b = 0; c = -25;
Δ = b2-4ac
Δ = 02-4·0.20·(-25)
Δ = 20
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:

$\sqrt{\Delta}=\sqrt{20}=\sqrt{4*5}=\sqrt{4}*\sqrt{5}=2\sqrt{5}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{5}}{2*0.20}=\frac{0-2\sqrt{5}}{0.4}
=-\frac{2\sqrt{5}}{0.4}
=-\frac{\sqrt{5}}{0.2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{5}}{2*0.20}=\frac{0+2\sqrt{5}}{0.4}
=\frac{2\sqrt{5}}{0.4}
=\frac{\sqrt{5}}{0.2}
$