0=-16t^2+18t+310

Solution for 0=-16t^2+18t+310 equation:

0=-16t^2+18t+310

We move all terms to the left:

0-(-16t^2+18t+310)=0

We add all the numbers together, and all the variables

-(-16t^2+18t+310)=0

We get rid of parentheses

16t^2-18t-310=0

a = 16; b = -18; c = -310;
Δ = b2-4ac
Δ = -182-4·16·(-310)
Δ = 20164
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{20164}=142$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-142}{2*16}=\frac{-124}{32}
=-3+7/8
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+142}{2*16}=\frac{160}{32}
=5
$