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-0.5x^2+40x-600=0
a = -0.5; b = 40; c = -600;
Δ = b2-4ac
Δ = 402-4·(-0.5)·(-600)
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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-20}{2*-0.5}=\frac{-60}{-1} =+60 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+20}{2*-0.5}=\frac{-20}{-1} =+20 $
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