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3x^2+13x-38=0
a = 3; b = 13; c = -38;
Δ = b2-4ac
Δ = 132-4·3·(-38)
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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-25}{2*3}=\frac{-38}{6} =-6+1/3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+25}{2*3}=\frac{12}{6} =2 $
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