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3x^2+19x-48=0
a = 3; b = 19; c = -48;
Δ = b2-4ac
Δ = 192-4·3·(-48)
Δ = 937
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{-(19)-\sqrt{937}}{2*3}=\frac{-19-\sqrt{937}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+\sqrt{937}}{2*3}=\frac{-19+\sqrt{937}}{6} $
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