(t)=-16t^2+160t+30

Solution for (t)=-16t^2+160t+30 equation:

(t)=-16t^2+160t+30

We move all terms to the left:

(t)-(-16t^2+160t+30)=0

We get rid of parentheses

16t^2-160t+t-30=0

We add all the numbers together, and all the variables

16t^2-159t-30=0

a = 16; b = -159; c = -30;
Δ = b2-4ac
Δ = -1592-4·16·(-30)
Δ = 27201
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{-(-159)-\sqrt{27201}}{2*16}=\frac{159-\sqrt{27201}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-159)+\sqrt{27201}}{2*16}=\frac{159+\sqrt{27201}}{32}
$