-t^2+6t+6=0

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

-t^2+6t+6=0

We add all the numbers together, and all the variables

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

a = -1; b = 6; c = +6;
Δ = b2-4ac
Δ = 62-4·(-1)·6
Δ = 60
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:

$\sqrt{\Delta}=\sqrt{60}=\sqrt{4*15}=\sqrt{4}*\sqrt{15}=2\sqrt{15}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{15}}{2*-1}=\frac{-6-2\sqrt{15}}{-2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{15}}{2*-1}=\frac{-6+2\sqrt{15}}{-2}
$