t=-0.15t^2+0.6t+1.8

Solution for t=-0.15t^2+0.6t+1.8 equation:

t=-0.15t^2+0.6t+1.8

We move all terms to the left:

t-(-0.15t^2+0.6t+1.8)=0

We get rid of parentheses

0.15t^2-0.6t+t-1.8=0

We add all the numbers together, and all the variables

0.15t^2+0.4t-1.8=0

a = 0.15; b = 0.4; c = -1.8;
Δ = b2-4ac
Δ = 0.42-4·0.15·(-1.8)
Δ = 1.24
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{-(0.4)-\sqrt{1.24}}{2*0.15}=\frac{-0.4-\sqrt{1.24}}{0.3}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0.4)+\sqrt{1.24}}{2*0.15}=\frac{-0.4+\sqrt{1.24}}{0.3}
$