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18x^2+29x=5
We move all terms to the left:
18x^2+29x-(5)=0
a = 18; b = 29; c = -5;
Δ = b2-4ac
Δ = 292-4·18·(-5)
Δ = 1201
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{-(29)-\sqrt{1201}}{2*18}=\frac{-29-\sqrt{1201}}{36} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(29)+\sqrt{1201}}{2*18}=\frac{-29+\sqrt{1201}}{36} $
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