100(t)=-16t^2+48t

Solution for 100(t)=-16t^2+48t equation:

100(t)=-16t^2+48t

We move all terms to the left:

100(t)-(-16t^2+48t)=0

We get rid of parentheses

16t^2-48t+100t=0

We add all the numbers together, and all the variables

16t^2+52t=0

a = 16; b = 52; c = 0;
Δ = b2-4ac
Δ = 522-4·16·0
Δ = 2704
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2704}=52$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(52)-52}{2*16}=\frac{-104}{32}
=-3+1/4
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(52)+52}{2*16}=\frac{0}{32}
=0
$