100=-16t^2+112t+5

Solution for 100=-16t^2+112t+5 equation:

100=-16t^2+112t+5

We move all terms to the left:

100-(-16t^2+112t+5)=0

We get rid of parentheses

16t^2-112t-5+100=0

We add all the numbers together, and all the variables

16t^2-112t+95=0

a = 16; b = -112; c = +95;
Δ = b2-4ac
Δ = -1122-4·16·95
Δ = 6464
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:

$\sqrt{\Delta}=\sqrt{6464}=\sqrt{64*101}=\sqrt{64}*\sqrt{101}=8\sqrt{101}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-112)-8\sqrt{101}}{2*16}=\frac{112-8\sqrt{101}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-112)+8\sqrt{101}}{2*16}=\frac{112+8\sqrt{101}}{32}
$