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33x^2+19x-2=0
a = 33; b = 19; c = -2;
Δ = b2-4ac
Δ = 192-4·33·(-2)
Δ = 625
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{625}=25$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-25}{2*33}=\frac{-44}{66} =-2/3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+25}{2*33}=\frac{6}{66} =1/11 $
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