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18x^2+39x+18=0
a = 18; b = 39; c = +18;
Δ = b2-4ac
Δ = 392-4·18·18
Δ = 225
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{225}=15$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(39)-15}{2*18}=\frac{-54}{36} =-1+1/2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(39)+15}{2*18}=\frac{-24}{36} =-2/3 $
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