-9.8t^2+20t+60=0

Solution for -9.8t^2+20t+60=0 equation:

-9.8t^2+20t+60=0

a = -9.8; b = 20; c = +60;
Δ = b2-4ac
Δ = 202-4·(-9.8)·60
Δ = 2752
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:

$\sqrt{\Delta}=\sqrt{2752}=\sqrt{64*43}=\sqrt{64}*\sqrt{43}=8\sqrt{43}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-8\sqrt{43}}{2*-9.8}=\frac{-20-8\sqrt{43}}{-19.6}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+8\sqrt{43}}{2*-9.8}=\frac{-20+8\sqrt{43}}{-19.6}
$