-0.25t^2+1.7t-2.5=0

Solution for -0.25t^2+1.7t-2.5=0 equation:

-0.25t^2+1.7t-2.5=0

a = -0.25; b = 1.7; c = -2.5;
Δ = b2-4ac
Δ = 1.72-4·(-0.25)·(-2.5)
Δ = 0.39
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{-(1.7)-\sqrt{0.39}}{2*-0.25}=\frac{-1.7-\sqrt{0.39}}{-0.5}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1.7)+\sqrt{0.39}}{2*-0.25}=\frac{-1.7+\sqrt{0.39}}{-0.5}
$