16=-16t^2+179t+0

Solution for 16=-16t^2+179t+0 equation:

16=-16t^2+179t+0

We move all terms to the left:

16-(-16t^2+179t+0)=0

We get rid of parentheses

16t^2-179t-0+16=0

We add all the numbers together, and all the variables

16t^2-179t+16=0

a = 16; b = -179; c = +16;
Δ = b2-4ac
Δ = -1792-4·16·16
Δ = 31017
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:

$\sqrt{\Delta}=\sqrt{31017}=\sqrt{49*633}=\sqrt{49}*\sqrt{633}=7\sqrt{633}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-179)-7\sqrt{633}}{2*16}=\frac{179-7\sqrt{633}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-179)+7\sqrt{633}}{2*16}=\frac{179+7\sqrt{633}}{32}
$