0=-16t^2+45t+35

Solution for 0=-16t^2+45t+35 equation:

0=-16t^2+45t+35

We move all terms to the left:

0-(-16t^2+45t+35)=0

We add all the numbers together, and all the variables

-(-16t^2+45t+35)=0

We get rid of parentheses

16t^2-45t-35=0

a = 16; b = -45; c = -35;
Δ = b2-4ac
Δ = -452-4·16·(-35)
Δ = 4265
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{-(-45)-\sqrt{4265}}{2*16}=\frac{45-\sqrt{4265}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-45)+\sqrt{4265}}{2*16}=\frac{45+\sqrt{4265}}{32}
$