40t-4.9t^2+18=0

Solution for 40t-4.9t^2+18=0 equation:

40t-4.9t^2+18=0

a = -4.9; b = 40; c = +18;
Δ = b2-4ac
Δ = 402-4·(-4.9)·18
Δ = 1952.8
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{-(40)-\sqrt{1952.8}}{2*-4.9}=\frac{-40-\sqrt{1952.8}}{-9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+\sqrt{1952.8}}{2*-4.9}=\frac{-40+\sqrt{1952.8}}{-9.8}
$