0=188-16t^2

Solution for 0=188-16t^2 equation:

0=188-16t^2

We move all terms to the left:

0-(188-16t^2)=0

We add all the numbers together, and all the variables

-(188-16t^2)=0

We get rid of parentheses

16t^2-188=0

a = 16; b = 0; c = -188;
Δ = b2-4ac
Δ = 02-4·16·(-188)
Δ = 12032
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:

$\sqrt{\Delta}=\sqrt{12032}=\sqrt{256*47}=\sqrt{256}*\sqrt{47}=16\sqrt{47}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16\sqrt{47}}{2*16}=\frac{0-16\sqrt{47}}{32}
=-\frac{16\sqrt{47}}{32}
=-\frac{\sqrt{47}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16\sqrt{47}}{2*16}=\frac{0+16\sqrt{47}}{32}
=\frac{16\sqrt{47}}{32}
=\frac{\sqrt{47}}{2}
$