0=-2t^2+48t+60

Solution for 0=-2t^2+48t+60 equation:

0=-2t^2+48t+60

We move all terms to the left:

0-(-2t^2+48t+60)=0

We add all the numbers together, and all the variables

-(-2t^2+48t+60)=0

We get rid of parentheses

2t^2-48t-60=0

a = 2; b = -48; c = -60;
Δ = b2-4ac
Δ = -482-4·2·(-60)
Δ = 2784
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:

$\sqrt{\Delta}=\sqrt{2784}=\sqrt{16*174}=\sqrt{16}*\sqrt{174}=4\sqrt{174}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-48)-4\sqrt{174}}{2*2}=\frac{48-4\sqrt{174}}{4}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-48)+4\sqrt{174}}{2*2}=\frac{48+4\sqrt{174}}{4}
$