2t^2-29t+14=0

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

2t^2-29t+14=0

a = 2; b = -29; c = +14;
Δ = b2-4ac
Δ = -292-4·2·14
Δ = 729
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{729}=27$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-29)-27}{2*2}=\frac{2}{4}
=1/2
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-29)+27}{2*2}=\frac{56}{4}
=14
$