45=-0.4t^2+12t

Solution for 45=-0.4t^2+12t equation:

45=-0.4t^2+12t

We move all terms to the left:

45-(-0.4t^2+12t)=0

We get rid of parentheses

0.4t^2-12t+45=0

a = 0.4; b = -12; c = +45;
Δ = b2-4ac
Δ = -122-4·0.4·45
Δ = 72
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:

$\sqrt{\Delta}=\sqrt{72}=\sqrt{36*2}=\sqrt{36}*\sqrt{2}=6\sqrt{2}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-6\sqrt{2}}{2*0.4}=\frac{12-6\sqrt{2}}{0.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+6\sqrt{2}}{2*0.4}=\frac{12+6\sqrt{2}}{0.8}
$