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18x^2-29x+3=0
a = 18; b = -29; c = +3;
Δ = b2-4ac
Δ = -292-4·18·3
Δ = 625
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{625}=25$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-29)-25}{2*18}=\frac{4}{36} =1/9 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-29)+25}{2*18}=\frac{54}{36} =1+1/2 $
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