.4t2=108

Solution for .4t2=108 equation:

.4t^2=108

We move all terms to the left:

.4t^2-(108)=0

a = .4; b = 0; c = -108;
Δ = b2-4ac
Δ = 02-4·.4·(-108)
Δ = 172.8
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{172.8}}{2*.4}=\frac{0-\sqrt{172.8}}{0.8}
=-\frac{\sqrt{}}{0.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+\sqrt{172.8}}{2*.4}=\frac{0+\sqrt{172.8}}{0.8}
=\frac{\sqrt{}}{0.8}
$