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3x^2+6x-5040=0
a = 3; b = 6; c = -5040;
Δ = b2-4ac
Δ = 62-4·3·(-5040)
Δ = 60516
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{60516}=246$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-246}{2*3}=\frac{-252}{6} =-42 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+246}{2*3}=\frac{240}{6} =40 $
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