-16t^2+16t+60=0

Solution for -16t^2+16t+60=0 equation:

-16t^2+16t+60=0

a = -16; b = 16; c = +60;
Δ = b2-4ac
Δ = 162-4·(-16)·60
Δ = 4096
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4096}=64$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-64}{2*-16}=\frac{-80}{-32}
=2+1/2
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+64}{2*-16}=\frac{48}{-32}
=-1+1/2
$