0.40t^2=1

Solution for 0.40t^2=1 equation:

0.40t^2=1

We move all terms to the left:

0.40t^2-(1)=0

a = 0.40; b = 0; c = -1;
Δ = b2-4ac
Δ = 02-4·0.40·(-1)
Δ = 1.6
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{-(0)-\sqrt{1.6}}{2*0.40}=\frac{0-\sqrt{1.6}}{0.8}
=-\frac{\sqrt{}}{0.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+\sqrt{1.6}}{2*0.40}=\frac{0+\sqrt{1.6}}{0.8}
=\frac{\sqrt{}}{0.8}
$