11=3+23t-16t^2

Solution for 11=3+23t-16t^2 equation:

11=3+23t-16t^2

We move all terms to the left:

11-(3+23t-16t^2)=0

We get rid of parentheses

16t^2-23t-3+11=0

We add all the numbers together, and all the variables

16t^2-23t+8=0

a = 16; b = -23; c = +8;
Δ = b2-4ac
Δ = -232-4·16·8
Δ = 17
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{-(-23)-\sqrt{17}}{2*16}=\frac{23-\sqrt{17}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-23)+\sqrt{17}}{2*16}=\frac{23+\sqrt{17}}{32}
$