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

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

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

a = .19; b = -5; c = +17;
Δ = b2-4ac
Δ = -52-4·.19·17
Δ = 12.08
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{12.08}}{2*.19}=\frac{5-\sqrt{12.08}}{0.38}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+\sqrt{12.08}}{2*.19}=\frac{5+\sqrt{12.08}}{0.38}
$