19t-1.86t^2=0

Solution for 19t-1.86t^2=0 equation:

19t-1.86t^2=0

a = -1.86; b = 19; c = 0;
Δ = b2-4ac
Δ = 192-4·(-1.86)·0
Δ = 361
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{361}=19$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-19}{2*-1.86}=\frac{-38}{-3.72}
=10+0.8/3.72
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+19}{2*-1.86}=\frac{0}{-3.72}
=0
$