0=16t^2-48t-18

Solution for 0=16t^2-48t-18 equation:

0=16t^2-48t-18

We move all terms to the left:

0-(16t^2-48t-18)=0

We add all the numbers together, and all the variables

-(16t^2-48t-18)=0

We get rid of parentheses

-16t^2+48t+18=0

a = -16; b = 48; c = +18;
Δ = b2-4ac
Δ = 482-4·(-16)·18
Δ = 3456
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:

$\sqrt{\Delta}=\sqrt{3456}=\sqrt{576*6}=\sqrt{576}*\sqrt{6}=24\sqrt{6}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(48)-24\sqrt{6}}{2*-16}=\frac{-48-24\sqrt{6}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(48)+24\sqrt{6}}{2*-16}=\frac{-48+24\sqrt{6}}{-32}
$