0=-16t^2+40t-7

Solution for 0=-16t^2+40t-7 equation:

0=-16t^2+40t-7

We move all terms to the left:

0-(-16t^2+40t-7)=0

We add all the numbers together, and all the variables

-(-16t^2+40t-7)=0

We get rid of parentheses

16t^2-40t+7=0

a = 16; b = -40; c = +7;
Δ = b2-4ac
Δ = -402-4·16·7
Δ = 1152
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1152}=\sqrt{576*2}=\sqrt{576}*\sqrt{2}=24\sqrt{2}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-24\sqrt{2}}{2*16}=\frac{40-24\sqrt{2}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+24\sqrt{2}}{2*16}=\frac{40+24\sqrt{2}}{32}
$