1.8t^2-10t+5=0

Solution for 1.8t^2-10t+5=0 equation:

1.8t^2-10t+5=0

a = 1.8; b = -10; c = +5;
Δ = b2-4ac
Δ = -102-4·1.8·5
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-8}{2*1.8}=\frac{2}{3.6}
=2/3.6
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+8}{2*1.8}=\frac{18}{3.6}
=5
$