0=-16t^2+16t+23

Solution for 0=-16t^2+16t+23 equation:

0=-16t^2+16t+23

We move all terms to the left:

0-(-16t^2+16t+23)=0

We add all the numbers together, and all the variables

-(-16t^2+16t+23)=0

We get rid of parentheses

16t^2-16t-23=0

a = 16; b = -16; c = -23;
Δ = b2-4ac
Δ = -162-4·16·(-23)
Δ = 1728
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:

$\sqrt{\Delta}=\sqrt{1728}=\sqrt{576*3}=\sqrt{576}*\sqrt{3}=24\sqrt{3}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-24\sqrt{3}}{2*16}=\frac{16-24\sqrt{3}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+24\sqrt{3}}{2*16}=\frac{16+24\sqrt{3}}{32}
$