t=-16t^2+40t+18

Solution for t=-16t^2+40t+18 equation:

t=-16t^2+40t+18

We move all terms to the left:

t-(-16t^2+40t+18)=0

We get rid of parentheses

16t^2-40t+t-18=0

We add all the numbers together, and all the variables

16t^2-39t-18=0

a = 16; b = -39; c = -18;
Δ = b2-4ac
Δ = -392-4·16·(-18)
Δ = 2673
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:

$\sqrt{\Delta}=\sqrt{2673}=\sqrt{81*33}=\sqrt{81}*\sqrt{33}=9\sqrt{33}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-39)-9\sqrt{33}}{2*16}=\frac{39-9\sqrt{33}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-39)+9\sqrt{33}}{2*16}=\frac{39+9\sqrt{33}}{32}
$