-.19t^2-5t+17=0

Solution for -.19t^2-5t+17=0 equation:

-.19t^2-5t+17=0

We add all the numbers together, and all the variables

-0.19t^2-5t+17=0

a = -0.19; b = -5; c = +17;
Δ = b2-4ac
Δ = -52-4·(-0.19)·17
Δ = 37.92
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{-(-5)-\sqrt{37.92}}{2*-0.19}=\frac{5-\sqrt{37.92}}{-0.38}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+\sqrt{37.92}}{2*-0.19}=\frac{5+\sqrt{37.92}}{-0.38}
$