14=16t^2+20t+6

Solution for 14=16t^2+20t+6 equation:

14=16t^2+20t+6

We move all terms to the left:

14-(16t^2+20t+6)=0

We get rid of parentheses

-16t^2-20t-6+14=0

We add all the numbers together, and all the variables

-16t^2-20t+8=0

a = -16; b = -20; c = +8;
Δ = b2-4ac
Δ = -202-4·(-16)·8
Δ = 912
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:

$\sqrt{\Delta}=\sqrt{912}=\sqrt{16*57}=\sqrt{16}*\sqrt{57}=4\sqrt{57}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-4\sqrt{57}}{2*-16}=\frac{20-4\sqrt{57}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+4\sqrt{57}}{2*-16}=\frac{20+4\sqrt{57}}{-32}
$