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

Solution for -0.4t^2+12t-45=0 equation:

-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}
$