10=-2t^2+48t+60

Solution for 10=-2t^2+48t+60 equation:

10=-2t^2+48t+60

We move all terms to the left:

10-(-2t^2+48t+60)=0

We get rid of parentheses

2t^2-48t-60+10=0

We add all the numbers together, and all the variables

2t^2-48t-50=0

a = 2; b = -48; c = -50;
Δ = b2-4ac
Δ = -482-4·2·(-50)
Δ = 2704
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2704}=52$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-48)-52}{2*2}=\frac{-4}{4}
=-1
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-48)+52}{2*2}=\frac{100}{4}
=25
$