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18x^2-3x-1=0
a = 18; b = -3; c = -1;
Δ = b2-4ac
Δ = -32-4·18·(-1)
Δ = 81
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{81}=9$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-9}{2*18}=\frac{-6}{36} =-1/6 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+9}{2*18}=\frac{12}{36} =1/3 $
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