3(3^2y-6)=27

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Solution for 3(3^2y-6)=27 equation:



3(3^2y-6)=27
We move all terms to the left:
3(3^2y-6)-(27)=0
We multiply parentheses
9y^2-18-27=0
We add all the numbers together, and all the variables
9y^2-45=0
a = 9; b = 0; c = -45;
Δ = b2-4ac
Δ = 02-4·9·(-45)
Δ = 1620
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:
$\sqrt{\Delta}=\sqrt{1620}=\sqrt{324*5}=\sqrt{324}*\sqrt{5}=18\sqrt{5}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-18\sqrt{5}}{2*9}=\frac{0-18\sqrt{5}}{18} =-\frac{18\sqrt{5}}{18} =-\sqrt{5} $
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+18\sqrt{5}}{2*9}=\frac{0+18\sqrt{5}}{18} =\frac{18\sqrt{5}}{18} =\sqrt{5} $

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