54(t)=-16t^2+104t+12

Solution for 54(t)=-16t^2+104t+12 equation:

54(t)=-16t^2+104t+12

We move all terms to the left:

54(t)-(-16t^2+104t+12)=0

We get rid of parentheses

16t^2-104t+54t-12=0

We add all the numbers together, and all the variables

16t^2-50t-12=0

a = 16; b = -50; c = -12;
Δ = b2-4ac
Δ = -502-4·16·(-12)
Δ = 3268
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:

$\sqrt{\Delta}=\sqrt{3268}=\sqrt{4*817}=\sqrt{4}*\sqrt{817}=2\sqrt{817}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-50)-2\sqrt{817}}{2*16}=\frac{50-2\sqrt{817}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-50)+2\sqrt{817}}{2*16}=\frac{50+2\sqrt{817}}{32}
$