27+12t-3t^2=0

Solution for 27+12t-3t^2=0 equation:

27+12t-3t^2=0

a = -3; b = 12; c = +27;
Δ = b2-4ac
Δ = 122-4·(-3)·27
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{-(12)-6\sqrt{13}}{2*-3}=\frac{-12-6\sqrt{13}}{-6}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+6\sqrt{13}}{2*-3}=\frac{-12+6\sqrt{13}}{-6}
$