1.9t^2+6t-40=0

Solution for 1.9t^2+6t-40=0 equation:

1.9t^2+6t-40=0

a = 1.9; b = 6; c = -40;
Δ = b2-4ac
Δ = 62-4·1.9·(-40)
Δ = 340
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{340}=\sqrt{4*85}=\sqrt{4}*\sqrt{85}=2\sqrt{85}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{85}}{2*1.9}=\frac{-6-2\sqrt{85}}{3.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{85}}{2*1.9}=\frac{-6+2\sqrt{85}}{3.8}
$