-4.9t^2+5t+40=0

Solution for -4.9t^2+5t+40=0 equation:

-4.9t^2+5t+40=0

a = -4.9; b = 5; c = +40;
Δ = b2-4ac
Δ = 52-4·(-4.9)·40
Δ = 809
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{-(5)-\sqrt{809}}{2*-4.9}=\frac{-5-\sqrt{809}}{-9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+\sqrt{809}}{2*-4.9}=\frac{-5+\sqrt{809}}{-9.8}
$