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20x^2+18x=18
We move all terms to the left:
20x^2+18x-(18)=0
a = 20; b = 18; c = -18;
Δ = b2-4ac
Δ = 182-4·20·(-18)
Δ = 1764
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{1764}=42$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-42}{2*20}=\frac{-60}{40} =-1+1/2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+42}{2*20}=\frac{24}{40} =3/5 $
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