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-2x(3x-9)=12
We move all terms to the left:
-2x(3x-9)-(12)=0
We multiply parentheses
-6x^2+18x-12=0
a = -6; b = 18; c = -12;
Δ = b2-4ac
Δ = 182-4·(-6)·(-12)
Δ = 36
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{36}=6$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-6}{2*-6}=\frac{-24}{-12} =+2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+6}{2*-6}=\frac{-12}{-12} =1 $
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