9t^2+18t+6=0

Solution for 9t^2+18t+6=0 equation:

9t^2+18t+6=0

a = 9; b = 18; c = +6;
Δ = b2-4ac
Δ = 182-4·9·6
Δ = 108
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:

$\sqrt{\Delta}=\sqrt{108}=\sqrt{36*3}=\sqrt{36}*\sqrt{3}=6\sqrt{3}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-6\sqrt{3}}{2*9}=\frac{-18-6\sqrt{3}}{18}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+6\sqrt{3}}{2*9}=\frac{-18+6\sqrt{3}}{18}
$