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18x^2-3x=6
We move all terms to the left:
18x^2-3x-(6)=0
a = 18; b = -3; c = -6;
Δ = b2-4ac
Δ = -32-4·18·(-6)
Δ = 441
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{441}=21$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-21}{2*18}=\frac{-18}{36} =-1/2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+21}{2*18}=\frac{24}{36} =2/3 $
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