1.86t^2+10t=0

Solution for 1.86t^2+10t=0 equation:

1.86t^2+10t=0

a = 1.86; b = 10; c = 0;
Δ = b2-4ac
Δ = 102-4·1.86·0
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-10}{2*1.86}=\frac{-20}{3.72}
=-5+1/2.6571428571429
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+10}{2*1.86}=\frac{0}{3.72}
=0
$