1.5t^2+4.9t+2=0

Solution for 1.5t^2+4.9t+2=0 equation:

1.5t^2+4.9t+2=0

a = 1.5; b = 4.9; c = +2;
Δ = b2-4ac
Δ = 4.92-4·1.5·2
Δ = 12.01
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{-(4.9)-\sqrt{12.01}}{2*1.5}=\frac{-4.9-\sqrt{12.01}}{3}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4.9)+\sqrt{12.01}}{2*1.5}=\frac{-4.9+\sqrt{12.01}}{3}
$