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