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3x^2+23x-18=0
a = 3; b = 23; c = -18;
Δ = b2-4ac
Δ = 232-4·3·(-18)
Δ = 745
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(23)-\sqrt{745}}{2*3}=\frac{-23-\sqrt{745}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+\sqrt{745}}{2*3}=\frac{-23+\sqrt{745}}{6} $
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