(3x-1)(3x-3)=4=

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Solution for (3x-1)(3x-3)=4= equation:



(3x-1)(3x-3)=4=
We move all terms to the left:
(3x-1)(3x-3)-(4)=0
We multiply parentheses ..
(+9x^2-9x-3x+3)-4=0
We get rid of parentheses
9x^2-9x-3x+3-4=0
We add all the numbers together, and all the variables
9x^2-12x-1=0
a = 9; b = -12; c = -1;
Δ = b2-4ac
Δ = -122-4·9·(-1)
Δ = 180
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:
$\sqrt{\Delta}=\sqrt{180}=\sqrt{36*5}=\sqrt{36}*\sqrt{5}=6\sqrt{5}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-6\sqrt{5}}{2*9}=\frac{12-6\sqrt{5}}{18} $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+6\sqrt{5}}{2*9}=\frac{12+6\sqrt{5}}{18} $

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