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18x^2+12x-6=0
a = 18; b = 12; c = -6;
Δ = b2-4ac
Δ = 122-4·18·(-6)
Δ = 576
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{576}=24$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-24}{2*18}=\frac{-36}{36} =-1 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+24}{2*18}=\frac{12}{36} =1/3 $
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