100=-16t^2+32+228

Solution for 100=-16t^2+32+228 equation:

100=-16t^2+32+228

We move all terms to the left:

100-(-16t^2+32+228)=0

We get rid of parentheses

16t^2-32-228+100=0

We add all the numbers together, and all the variables

16t^2-160=0

a = 16; b = 0; c = -160;
Δ = b2-4ac
Δ = 02-4·16·(-160)
Δ = 10240
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:

$\sqrt{\Delta}=\sqrt{10240}=\sqrt{1024*10}=\sqrt{1024}*\sqrt{10}=32\sqrt{10}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-32\sqrt{10}}{2*16}=\frac{0-32\sqrt{10}}{32}
=-\frac{32\sqrt{10}}{32}
=-\sqrt{10}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+32\sqrt{10}}{2*16}=\frac{0+32\sqrt{10}}{32}
=\frac{32\sqrt{10}}{32}
=\sqrt{10}
$