-2t^2+6t+1=0

Solution for -2t^2+6t+1=0 equation:

-2t^2+6t+1=0

a = -2; b = 6; c = +1;
Δ = b2-4ac
Δ = 62-4·(-2)·1
Δ = 44
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:

$\sqrt{\Delta}=\sqrt{44}=\sqrt{4*11}=\sqrt{4}*\sqrt{11}=2\sqrt{11}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{11}}{2*-2}=\frac{-6-2\sqrt{11}}{-4}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{11}}{2*-2}=\frac{-6+2\sqrt{11}}{-4}
$