100=-4.9t^2+16t+300

Solution for 100=-4.9t^2+16t+300 equation:

100=-4.9t^2+16t+300

We move all terms to the left:

100-(-4.9t^2+16t+300)=0

We get rid of parentheses

4.9t^2-16t-300+100=0

We add all the numbers together, and all the variables

4.9t^2-16t-200=0

a = 4.9; b = -16; c = -200;
Δ = b2-4ac
Δ = -162-4·4.9·(-200)
Δ = 4176
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:

$\sqrt{\Delta}=\sqrt{4176}=\sqrt{144*29}=\sqrt{144}*\sqrt{29}=12\sqrt{29}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-12\sqrt{29}}{2*4.9}=\frac{16-12\sqrt{29}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+12\sqrt{29}}{2*4.9}=\frac{16+12\sqrt{29}}{9.8}
$