0=16t^2-48t-36

Solution for 0=16t^2-48t-36 equation:

0=16t^2-48t-36

We move all terms to the left:

0-(16t^2-48t-36)=0

We add all the numbers together, and all the variables

-(16t^2-48t-36)=0

We get rid of parentheses

-16t^2+48t+36=0

a = -16; b = 48; c = +36;
Δ = b2-4ac
Δ = 482-4·(-16)·36
Δ = 4608
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:

$\sqrt{\Delta}=\sqrt{4608}=\sqrt{2304*2}=\sqrt{2304}*\sqrt{2}=48\sqrt{2}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(48)-48\sqrt{2}}{2*-16}=\frac{-48-48\sqrt{2}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(48)+48\sqrt{2}}{2*-16}=\frac{-48+48\sqrt{2}}{-32}
$