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 = 432
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:

$\sqrt{\Delta}=\sqrt{432}=\sqrt{144*3}=\sqrt{144}*\sqrt{3}=12\sqrt{3}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-12\sqrt{3}}{2*3}=\frac{18-12\sqrt{3}}{6}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+12\sqrt{3}}{2*3}=\frac{18+12\sqrt{3}}{6}
$