(t)=-16t^2+55t+4

Solution for (t)=-16t^2+55t+4 equation:

(t)=-16t^2+55t+4

We move all terms to the left:

(t)-(-16t^2+55t+4)=0

We get rid of parentheses

16t^2-55t+t-4=0

We add all the numbers together, and all the variables

16t^2-54t-4=0

a = 16; b = -54; c = -4;
Δ = b2-4ac
Δ = -542-4·16·(-4)
Δ = 3172
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:

$\sqrt{\Delta}=\sqrt{3172}=\sqrt{4*793}=\sqrt{4}*\sqrt{793}=2\sqrt{793}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-54)-2\sqrt{793}}{2*16}=\frac{54-2\sqrt{793}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-54)+2\sqrt{793}}{2*16}=\frac{54+2\sqrt{793}}{32}
$