(t)=-16t^2+30t+6

Solution for (t)=-16t^2+30t+6 equation:

(t)=-16t^2+30t+6

We move all terms to the left:

(t)-(-16t^2+30t+6)=0

We get rid of parentheses

16t^2-30t+t-6=0

We add all the numbers together, and all the variables

16t^2-29t-6=0

a = 16; b = -29; c = -6;
Δ = b2-4ac
Δ = -292-4·16·(-6)
Δ = 1225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1225}=35$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-29)-35}{2*16}=\frac{-6}{32}
=-3/16
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-29)+35}{2*16}=\frac{64}{32}
=2
$