3t^2+18t+9=0

Solution for 3t^2+18t+9=0 equation:

3t^2+18t+9=0

a = 3; b = 18; c = +9;
Δ = b2-4ac
Δ = 182-4·3·9
Δ = 216
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:

$\sqrt{\Delta}=\sqrt{216}=\sqrt{36*6}=\sqrt{36}*\sqrt{6}=6\sqrt{6}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-6\sqrt{6}}{2*3}=\frac{-18-6\sqrt{6}}{6}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+6\sqrt{6}}{2*3}=\frac{-18+6\sqrt{6}}{6}
$