2t^2+12t+3=0

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

2t^2+12t+3=0

a = 2; b = 12; c = +3;
Δ = b2-4ac
Δ = 122-4·2·3
Δ = 120
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:

$\sqrt{\Delta}=\sqrt{120}=\sqrt{4*30}=\sqrt{4}*\sqrt{30}=2\sqrt{30}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-2\sqrt{30}}{2*2}=\frac{-12-2\sqrt{30}}{4}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+2\sqrt{30}}{2*2}=\frac{-12+2\sqrt{30}}{4}
$