-2.7t^2+40t-7.5=0

Solution for -2.7t^2+40t-7.5=0 equation:

-2.7t^2+40t-7.5=0

a = -2.7; b = 40; c = -7.5;
Δ = b2-4ac
Δ = 402-4·(-2.7)·(-7.5)
Δ = 1519
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:

$\sqrt{\Delta}=\sqrt{1519}=\sqrt{49*31}=\sqrt{49}*\sqrt{31}=7\sqrt{31}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-7\sqrt{31}}{2*-2.7}=\frac{-40-7\sqrt{31}}{-5.4}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+7\sqrt{31}}{2*-2.7}=\frac{-40+7\sqrt{31}}{-5.4}
$