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18x^2+93x-120=0
a = 18; b = 93; c = -120;
Δ = b2-4ac
Δ = 932-4·18·(-120)
Δ = 17289
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{17289}=\sqrt{9*1921}=\sqrt{9}*\sqrt{1921}=3\sqrt{1921}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(93)-3\sqrt{1921}}{2*18}=\frac{-93-3\sqrt{1921}}{36} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(93)+3\sqrt{1921}}{2*18}=\frac{-93+3\sqrt{1921}}{36} $
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