60=16t-(t^2)

Solution for 60=16t-(t^2) equation:

60=16t-(t^2)

We move all terms to the left:

60-(16t-(t^2))=0

We get rid of parentheses

t^2-16t+60=0

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

$\sqrt{\Delta}=\sqrt{16}=4$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-4}{2*1}=\frac{12}{2}
=6
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+4}{2*1}=\frac{20}{2}
=10
$