14t^2-5t-6=0

Solution for 14t^2-5t-6=0 equation:

14t^2-5t-6=0

a = 14; b = -5; c = -6;
Δ = b2-4ac
Δ = -52-4·14·(-6)
Δ = 361
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{361}=19$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-19}{2*14}=\frac{-14}{28}
=-1/2
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+19}{2*14}=\frac{24}{28}
=6/7
$