0=-16t2+135t+56

Solution for 0=-16t2+135t+56 equation:

0=-16t^2+135t+56

We move all terms to the left:

0-(-16t^2+135t+56)=0

We add all the numbers together, and all the variables

-(-16t^2+135t+56)=0

We get rid of parentheses

16t^2-135t-56=0

a = 16; b = -135; c = -56;
Δ = b2-4ac
Δ = -1352-4·16·(-56)
Δ = 21809
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{-(-135)-\sqrt{21809}}{2*16}=\frac{135-\sqrt{21809}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-135)+\sqrt{21809}}{2*16}=\frac{135+\sqrt{21809}}{32}
$