2t^2-2t-6=0

Solution for 2t^2-2t-6=0 equation:

2t^2-2t-6=0

a = 2; b = -2; c = -6;
Δ = b2-4ac
Δ = -22-4·2·(-6)
Δ = 52
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:

$\sqrt{\Delta}=\sqrt{52}=\sqrt{4*13}=\sqrt{4}*\sqrt{13}=2\sqrt{13}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{13}}{2*2}=\frac{2-2\sqrt{13}}{4}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{13}}{2*2}=\frac{2+2\sqrt{13}}{4}
$