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3(2x-3)x=1
We move all terms to the left:
3(2x-3)x-(1)=0
We multiply parentheses
6x^2-9x-1=0
a = 6; b = -9; c = -1;
Δ = b2-4ac
Δ = -92-4·6·(-1)
Δ = 105
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-\sqrt{105}}{2*6}=\frac{9-\sqrt{105}}{12} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+\sqrt{105}}{2*6}=\frac{9+\sqrt{105}}{12} $
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