t^2+2.5t=-1.44

Solution for t^2+2.5t=-1.44 equation:

t^2+2.5t=-1.44

We move all terms to the left:

t^2+2.5t-(-1.44)=0

We add all the numbers together, and all the variables

t^2+2.5t+1.44=0

a = 1; b = 2.5; c = +1.44;
Δ = b2-4ac
Δ = 2.52-4·1·1.44
Δ = 0.49
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{-(2.5)-\sqrt{0.49}}{2*1}=\frac{-2.5-\sqrt{0.49}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2.5)+\sqrt{0.49}}{2*1}=\frac{-2.5+\sqrt{0.49}}{2}
$