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-0.5x^2+10x=0
a = -0.5; b = 10; c = 0;
Δ = b2-4ac
Δ = 102-4·(-0.5)·0
Δ = 100
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{100}=10$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-10}{2*-0.5}=\frac{-20}{-1} =+20 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+10}{2*-0.5}=\frac{0}{-1} =0 $
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