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-0.2x^2+40x-1500=0
a = -0.2; b = 40; c = -1500;
Δ = b2-4ac
Δ = 402-4·(-0.2)·(-1500)
Δ = 400
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{400}=20$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-20}{2*-0.2}=\frac{-60}{-0.4} =+150 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+20}{2*-0.2}=\frac{-20}{-0.4} =+50 $
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