0=-16t^2+95t+3.5

Solution for 0=-16t^2+95t+3.5 equation:

0=-16t^2+95t+3.5

We move all terms to the left:

0-(-16t^2+95t+3.5)=0

We add all the numbers together, and all the variables

-(-16t^2+95t+3.5)=0

We get rid of parentheses

16t^2-95t-3.5=0

a = 16; b = -95; c = -3.5;
Δ = b2-4ac
Δ = -952-4·16·(-3.5)
Δ = 9249
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{-(-95)-\sqrt{9249}}{2*16}=\frac{95-\sqrt{9249}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-95)+\sqrt{9249}}{2*16}=\frac{95+\sqrt{9249}}{32}
$