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2000+40x-3x^2=0
a = -3; b = 40; c = +2000;
Δ = b2-4ac
Δ = 402-4·(-3)·2000
Δ = 25600
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{25600}=160$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-160}{2*-3}=\frac{-200}{-6} =33+1/3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+160}{2*-3}=\frac{120}{-6} =-20 $
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