0=230-16t^2

Solution for 0=230-16t^2 equation:

0=230-16t^2

We move all terms to the left:

0-(230-16t^2)=0

We add all the numbers together, and all the variables

-(230-16t^2)=0

We get rid of parentheses

16t^2-230=0

a = 16; b = 0; c = -230;
Δ = b2-4ac
Δ = 02-4·16·(-230)
Δ = 14720
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:

$\sqrt{\Delta}=\sqrt{14720}=\sqrt{64*230}=\sqrt{64}*\sqrt{230}=8\sqrt{230}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{230}}{2*16}=\frac{0-8\sqrt{230}}{32}
=-\frac{8\sqrt{230}}{32}
=-\frac{\sqrt{230}}{4}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{230}}{2*16}=\frac{0+8\sqrt{230}}{32}
=\frac{8\sqrt{230}}{32}
=\frac{\sqrt{230}}{4}
$