30+40t-4.9t^2=0

Solution for 30+40t-4.9t^2=0 equation:

30+40t-4.9t^2=0

a = -4.9; b = 40; c = +30;
Δ = b2-4ac
Δ = 402-4·(-4.9)·30
Δ = 2188
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:

$\sqrt{\Delta}=\sqrt{2188}=\sqrt{4*547}=\sqrt{4}*\sqrt{547}=2\sqrt{547}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-2\sqrt{547}}{2*-4.9}=\frac{-40-2\sqrt{547}}{-9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+2\sqrt{547}}{2*-4.9}=\frac{-40+2\sqrt{547}}{-9.8}
$