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18x^2+93x+120=0
a = 18; b = 93; c = +120;
Δ = b2-4ac
Δ = 932-4·18·120
Δ = 9
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{9}=3$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(93)-3}{2*18}=\frac{-96}{36} =-2+2/3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(93)+3}{2*18}=\frac{-90}{36} =-2+1/2 $
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