3t^2+12t-2=0

Solution for 3t^2+12t-2=0 equation:

3t^2+12t-2=0

a = 3; b = 12; c = -2;
Δ = b2-4ac
Δ = 122-4·3·(-2)
Δ = 168
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:

$\sqrt{\Delta}=\sqrt{168}=\sqrt{4*42}=\sqrt{4}*\sqrt{42}=2\sqrt{42}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-2\sqrt{42}}{2*3}=\frac{-12-2\sqrt{42}}{6}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+2\sqrt{42}}{2*3}=\frac{-12+2\sqrt{42}}{6}
$