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/6(x+1)(x-1)=13x-12
We move all terms to the left:
/6(x+1)(x-1)-(13x-12)=0
Domain of the equation: 6(x+1)(x-1)!=0We use the square of the difference formula
x∈R
x^2-(13x-12)-1=0
We get rid of parentheses
x^2-13x+12-1=0
We add all the numbers together, and all the variables
x^2-13x+11=0
a = 1; b = -13; c = +11;
Δ = b2-4ac
Δ = -132-4·1·11
Δ = 125
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{125}=\sqrt{25*5}=\sqrt{25}*\sqrt{5}=5\sqrt{5}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-13)-5\sqrt{5}}{2*1}=\frac{13-5\sqrt{5}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+5\sqrt{5}}{2*1}=\frac{13+5\sqrt{5}}{2} $
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