(t)=14+32t-16t^2

Solution for (t)=14+32t-16t^2 equation:

(t)=14+32t-16t^2

We move all terms to the left:

(t)-(14+32t-16t^2)=0

We get rid of parentheses

16t^2-32t+t-14=0

We add all the numbers together, and all the variables

16t^2-31t-14=0

a = 16; b = -31; c = -14;
Δ = b2-4ac
Δ = -312-4·16·(-14)
Δ = 1857
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{-(-31)-\sqrt{1857}}{2*16}=\frac{31-\sqrt{1857}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-31)+\sqrt{1857}}{2*16}=\frac{31+\sqrt{1857}}{32}
$