0=-16t^2+95t+11

Solution for 0=-16t^2+95t+11 equation:

0=-16t^2+95t+11

We move all terms to the left:

0-(-16t^2+95t+11)=0

We add all the numbers together, and all the variables

-(-16t^2+95t+11)=0

We get rid of parentheses

16t^2-95t-11=0

a = 16; b = -95; c = -11;
Δ = b2-4ac
Δ = -952-4·16·(-11)
Δ = 9729
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:

$\sqrt{\Delta}=\sqrt{9729}=\sqrt{9*1081}=\sqrt{9}*\sqrt{1081}=3\sqrt{1081}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-95)-3\sqrt{1081}}{2*16}=\frac{95-3\sqrt{1081}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-95)+3\sqrt{1081}}{2*16}=\frac{95+3\sqrt{1081}}{32}
$