38=-16t^2+48+2

Solution for 38=-16t^2+48+2 equation:

38=-16t^2+48+2

We move all terms to the left:

38-(-16t^2+48+2)=0

We get rid of parentheses

16t^2-48-2+38=0

We add all the numbers together, and all the variables

16t^2-12=0

a = 16; b = 0; c = -12;
Δ = b2-4ac
Δ = 02-4·16·(-12)
Δ = 768
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:

$\sqrt{\Delta}=\sqrt{768}=\sqrt{256*3}=\sqrt{256}*\sqrt{3}=16\sqrt{3}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16\sqrt{3}}{2*16}=\frac{0-16\sqrt{3}}{32}
=-\frac{16\sqrt{3}}{32}
=-\frac{\sqrt{3}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16\sqrt{3}}{2*16}=\frac{0+16\sqrt{3}}{32}
=\frac{16\sqrt{3}}{32}
=\frac{\sqrt{3}}{2}
$