2t^2-12t-4=0

Solution for 2t^2-12t-4=0 equation:

2t^2-12t-4=0

a = 2; b = -12; c = -4;
Δ = b2-4ac
Δ = -122-4·2·(-4)
Δ = 176
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:

$\sqrt{\Delta}=\sqrt{176}=\sqrt{16*11}=\sqrt{16}*\sqrt{11}=4\sqrt{11}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-4\sqrt{11}}{2*2}=\frac{12-4\sqrt{11}}{4}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+4\sqrt{11}}{2*2}=\frac{12+4\sqrt{11}}{4}
$