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0.5x^2-16x+96=0
a = 0.5; b = -16; c = +96;
Δ = b2-4ac
Δ = -162-4·0.5·96
Δ = 64
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{64}=8$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-8}{2*0.5}=\frac{8}{1} =8 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+8}{2*0.5}=\frac{24}{1} =24 $
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