3(t)=-16t2+96t+112

Solution for 3(t)=-16t2+96t+112 equation:

3(t)=-16t^2+96t+112

We move all terms to the left:

3(t)-(-16t^2+96t+112)=0

We get rid of parentheses

16t^2-96t+3t-112=0

We add all the numbers together, and all the variables

16t^2-93t-112=0

a = 16; b = -93; c = -112;
Δ = b2-4ac
Δ = -932-4·16·(-112)
Δ = 15817
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-93)-\sqrt{15817}}{2*16}=\frac{93-\sqrt{15817}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-93)+\sqrt{15817}}{2*16}=\frac{93+\sqrt{15817}}{32}
$