100(t)=-16t^2+80t+32

Solution for 100(t)=-16t^2+80t+32 equation:

100(t)=-16t^2+80t+32

We move all terms to the left:

100(t)-(-16t^2+80t+32)=0

We get rid of parentheses

16t^2-80t+100t-32=0

We add all the numbers together, and all the variables

16t^2+20t-32=0

a = 16; b = 20; c = -32;
Δ = b2-4ac
Δ = 202-4·16·(-32)
Δ = 2448
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:

$\sqrt{\Delta}=\sqrt{2448}=\sqrt{144*17}=\sqrt{144}*\sqrt{17}=12\sqrt{17}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-12\sqrt{17}}{2*16}=\frac{-20-12\sqrt{17}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+12\sqrt{17}}{2*16}=\frac{-20+12\sqrt{17}}{32}
$