3t^2+24t+9=0

Solution for 3t^2+24t+9=0 equation:

3t^2+24t+9=0

a = 3; b = 24; c = +9;
Δ = b2-4ac
Δ = 242-4·3·9
Δ = 468
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:

$\sqrt{\Delta}=\sqrt{468}=\sqrt{36*13}=\sqrt{36}*\sqrt{13}=6\sqrt{13}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-6\sqrt{13}}{2*3}=\frac{-24-6\sqrt{13}}{6}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+6\sqrt{13}}{2*3}=\frac{-24+6\sqrt{13}}{6}
$