.4t^2=6t+18

Solution for .4t^2=6t+18 equation:

.4t^2=6t+18

We move all terms to the left:

.4t^2-(6t+18)=0

We get rid of parentheses

.4t^2-6t-18=0

a = .4; b = -6; c = -18;
Δ = b2-4ac
Δ = -62-4·.4·(-18)
Δ = 64.8
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{-(-6)-\sqrt{64.8}}{2*.4}=\frac{6-\sqrt{64.8}}{0.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+\sqrt{64.8}}{2*.4}=\frac{6+\sqrt{64.8}}{0.8}
$