0=16t^2+40t+9

Solution for 0=16t^2+40t+9 equation:

0=16t^2+40t+9

We move all terms to the left:

0-(16t^2+40t+9)=0

We add all the numbers together, and all the variables

-(16t^2+40t+9)=0

We get rid of parentheses

-16t^2-40t-9=0

a = -16; b = -40; c = -9;
Δ = b2-4ac
Δ = -402-4·(-16)·(-9)
Δ = 1024
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1024}=32$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-32}{2*-16}=\frac{8}{-32}
=-1/4
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+32}{2*-16}=\frac{72}{-32}
=-2+1/4
$