6t^2-27=0

Solution for 6t^2-27=0 equation:

6t^2-27=0

a = 6; b = 0; c = -27;
Δ = b2-4ac
Δ = 02-4·6·(-27)
Δ = 648
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:

$\sqrt{\Delta}=\sqrt{648}=\sqrt{324*2}=\sqrt{324}*\sqrt{2}=18\sqrt{2}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-18\sqrt{2}}{2*6}=\frac{0-18\sqrt{2}}{12}
=-\frac{18\sqrt{2}}{12}
=-\frac{3\sqrt{2}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+18\sqrt{2}}{2*6}=\frac{0+18\sqrt{2}}{12}
=\frac{18\sqrt{2}}{12}
=\frac{3\sqrt{2}}{2}
$