0=16t^2+38*8t+3

Solution for 0=16t^2+38*8t+3 equation:

0=16t^2+38*8t+3

We move all terms to the left:

0-(16t^2+38*8t+3)=0

We add all the numbers together, and all the variables

-(16t^2+38*8t+3)=0

We get rid of parentheses

-16t^2-38*8t-3=0

Wy multiply elements

-16t^2-304t-3=0

a = -16; b = -304; c = -3;
Δ = b2-4ac
Δ = -3042-4·(-16)·(-3)
Δ = 92224
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:

$\sqrt{\Delta}=\sqrt{92224}=\sqrt{64*1441}=\sqrt{64}*\sqrt{1441}=8\sqrt{1441}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-304)-8\sqrt{1441}}{2*-16}=\frac{304-8\sqrt{1441}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-304)+8\sqrt{1441}}{2*-16}=\frac{304+8\sqrt{1441}}{-32}
$