6t^2+20t+14=0

Solution for 6t^2+20t+14=0 equation:

6t^2+20t+14=0

a = 6; b = 20; c = +14;
Δ = b2-4ac
Δ = 202-4·6·14
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-8}{2*6}=\frac{-28}{12}
=-2+1/3
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+8}{2*6}=\frac{-12}{12}
=-1
$