3t^2+14t+5=0

Solution for 3t^2+14t+5=0 equation:

3t^2+14t+5=0

a = 3; b = 14; c = +5;
Δ = b2-4ac
Δ = 142-4·3·5
Δ = 136
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{136}=\sqrt{4*34}=\sqrt{4}*\sqrt{34}=2\sqrt{34}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-2\sqrt{34}}{2*3}=\frac{-14-2\sqrt{34}}{6}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+2\sqrt{34}}{2*3}=\frac{-14+2\sqrt{34}}{6}
$