-4.9t^2+60t+100=0

Solution for -4.9t^2+60t+100=0 equation:

-4.9t^2+60t+100=0

a = -4.9; b = 60; c = +100;
Δ = b2-4ac
Δ = 602-4·(-4.9)·100
Δ = 5560
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:

$\sqrt{\Delta}=\sqrt{5560}=\sqrt{4*1390}=\sqrt{4}*\sqrt{1390}=2\sqrt{1390}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(60)-2\sqrt{1390}}{2*-4.9}=\frac{-60-2\sqrt{1390}}{-9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(60)+2\sqrt{1390}}{2*-4.9}=\frac{-60+2\sqrt{1390}}{-9.8}
$