723=-16t^2+729

Solution for 723=-16t^2+729 equation:

723=-16t^2+729

We move all terms to the left:

723-(-16t^2+729)=0

We get rid of parentheses

16t^2-729+723=0

We add all the numbers together, and all the variables

16t^2-6=0

a = 16; b = 0; c = -6;
Δ = b2-4ac
Δ = 02-4·16·(-6)
Δ = 384
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:

$\sqrt{\Delta}=\sqrt{384}=\sqrt{64*6}=\sqrt{64}*\sqrt{6}=8\sqrt{6}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{6}}{2*16}=\frac{0-8\sqrt{6}}{32}
=-\frac{8\sqrt{6}}{32}
=-\frac{\sqrt{6}}{4}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{6}}{2*16}=\frac{0+8\sqrt{6}}{32}
=\frac{8\sqrt{6}}{32}
=\frac{\sqrt{6}}{4}
$