15t^2+29t-14=0

Solution for 15t^2+29t-14=0 equation:

15t^2+29t-14=0

a = 15; b = 29; c = -14;
Δ = b2-4ac
Δ = 292-4·15·(-14)
Δ = 1681
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1681}=41$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(29)-41}{2*15}=\frac{-70}{30}
=-2+1/3
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(29)+41}{2*15}=\frac{12}{30}
=2/5
$