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9x(3x-3)=27
We move all terms to the left:
9x(3x-3)-(27)=0
We multiply parentheses
27x^2-27x-27=0
a = 27; b = -27; c = -27;
Δ = b2-4ac
Δ = -272-4·27·(-27)
Δ = 3645
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{3645}=\sqrt{729*5}=\sqrt{729}*\sqrt{5}=27\sqrt{5}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-27)-27\sqrt{5}}{2*27}=\frac{27-27\sqrt{5}}{54} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-27)+27\sqrt{5}}{2*27}=\frac{27+27\sqrt{5}}{54} $
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