-16t^2+23t+6=14

Solution for -16t^2+23t+6=14 equation:

-16t^2+23t+6=14

We move all terms to the left:

-16t^2+23t+6-(14)=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}
$