0=38t^2+48t+2

Solution for 0=38t^2+48t+2 equation:

0=38t^2+48t+2

We move all terms to the left:

0-(38t^2+48t+2)=0

We add all the numbers together, and all the variables

-(38t^2+48t+2)=0

We get rid of parentheses

-38t^2-48t-2=0

a = -38; b = -48; c = -2;
Δ = b2-4ac
Δ = -482-4·(-38)·(-2)
Δ = 2000
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:

$\sqrt{\Delta}=\sqrt{2000}=\sqrt{400*5}=\sqrt{400}*\sqrt{5}=20\sqrt{5}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-48)-20\sqrt{5}}{2*-38}=\frac{48-20\sqrt{5}}{-76}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-48)+20\sqrt{5}}{2*-38}=\frac{48+20\sqrt{5}}{-76}
$