-4.9t^2+20t+60=0

Solution for -4.9t^2+20t+60=0 equation:

-4.9t^2+20t+60=0

a = -4.9; b = 20; c = +60;
Δ = b2-4ac
Δ = 202-4·(-4.9)·60
Δ = 1576
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:

$\sqrt{\Delta}=\sqrt{1576}=\sqrt{4*394}=\sqrt{4}*\sqrt{394}=2\sqrt{394}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-2\sqrt{394}}{2*-4.9}=\frac{-20-2\sqrt{394}}{-9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+2\sqrt{394}}{2*-4.9}=\frac{-20+2\sqrt{394}}{-9.8}
$