.0825t^2-0.928t+1.831=0

Solution for .0825t^2-0.928t+1.831=0 equation:

.0825t^2-0.928t+1.831=0

a = .0825; b = -0.928; c = +1.831;
Δ = b2-4ac
Δ = -0.9282-4·.0825·1.831
Δ = 0.256954
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{-(-0.928)-\sqrt{0.256954}}{2*.0825}=\frac{0.928-\sqrt{0.256954}}{0.165}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-0.928)+\sqrt{0.256954}}{2*.0825}=\frac{0.928+\sqrt{0.256954}}{0.165}
$