60=1t^2

Solution for 60=1t^2 equation:

60=1t^2

We move all terms to the left:

60-(1t^2)=0

a = -1; b = 0; c = +60;
Δ = b2-4ac
Δ = 02-4·(-1)·60
Δ = 240
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:

$\sqrt{\Delta}=\sqrt{240}=\sqrt{16*15}=\sqrt{16}*\sqrt{15}=4\sqrt{15}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{15}}{2*-1}=\frac{0-4\sqrt{15}}{-2}
=-\frac{4\sqrt{15}}{-2}
=-\frac{2\sqrt{15}}{-1}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{15}}{2*-1}=\frac{0+4\sqrt{15}}{-2}
=\frac{4\sqrt{15}}{-2}
=\frac{2\sqrt{15}}{-1}
$