(t)=-16t2+16t+480

Solution for (t)=-16t2+16t+480 equation:

(t)=-16t^2+16t+480

We move all terms to the left:

(t)-(-16t^2+16t+480)=0

We get rid of parentheses

16t^2-16t+t-480=0

We add all the numbers together, and all the variables

16t^2-15t-480=0

a = 16; b = -15; c = -480;
Δ = b2-4ac
Δ = -152-4·16·(-480)
Δ = 30945
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{-(-15)-\sqrt{30945}}{2*16}=\frac{15-\sqrt{30945}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+\sqrt{30945}}{2*16}=\frac{15+\sqrt{30945}}{32}
$