3t^2-324=0

Solution for 3t^2-324=0 equation:

3t^2-324=0

a = 3; b = 0; c = -324;
Δ = b2-4ac
Δ = 02-4·3·(-324)
Δ = 3888
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:

$\sqrt{\Delta}=\sqrt{3888}=\sqrt{1296*3}=\sqrt{1296}*\sqrt{3}=36\sqrt{3}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-36\sqrt{3}}{2*3}=\frac{0-36\sqrt{3}}{6}
=-\frac{36\sqrt{3}}{6}
=-6\sqrt{3}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+36\sqrt{3}}{2*3}=\frac{0+36\sqrt{3}}{6}
=\frac{36\sqrt{3}}{6}
=6\sqrt{3}
$