0(t)=-16t^2+40t+6

Solution for 0(t)=-16t^2+40t+6 equation:

0(t)=-16t^2+40t+6

We move all terms to the left:

0(t)-(-16t^2+40t+6)=0

We add all the numbers together, and all the variables

-(-16t^2+40t+6)+t=0

We get rid of parentheses

16t^2-40t+t-6=0

We add all the numbers together, and all the variables

16t^2-39t-6=0

a = 16; b = -39; c = -6;
Δ = b2-4ac
Δ = -392-4·16·(-6)
Δ = 1905
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{-(-39)-\sqrt{1905}}{2*16}=\frac{39-\sqrt{1905}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-39)+\sqrt{1905}}{2*16}=\frac{39+\sqrt{1905}}{32}
$