6(3^2t-1)+3=27

Solution for 6(3^2t-1)+3=27 equation:

6(3^2t-1)+3=27

We move all terms to the left:

6(3^2t-1)+3-(27)=0

We add all the numbers together, and all the variables

6(3^2t-1)-24=0

We multiply parentheses

18t^2-6-24=0

We add all the numbers together, and all the variables

18t^2-30=0

a = 18; b = 0; c = -30;
Δ = b2-4ac
Δ = 02-4·18·(-30)
Δ = 2160
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:

$\sqrt{\Delta}=\sqrt{2160}=\sqrt{144*15}=\sqrt{144}*\sqrt{15}=12\sqrt{15}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-12\sqrt{15}}{2*18}=\frac{0-12\sqrt{15}}{36}
=-\frac{12\sqrt{15}}{36}
=-\frac{\sqrt{15}}{3}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+12\sqrt{15}}{2*18}=\frac{0+12\sqrt{15}}{36}
=\frac{12\sqrt{15}}{36}
=\frac{\sqrt{15}}{3}
$