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0.5x^2-40x=-512
We move all terms to the left:
0.5x^2-40x-(-512)=0
We add all the numbers together, and all the variables
0.5x^2-40x+512=0
a = 0.5; b = -40; c = +512;
Δ = b2-4ac
Δ = -402-4·0.5·512
Δ = 576
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{576}=24$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-24}{2*0.5}=\frac{16}{1} =16 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+24}{2*0.5}=\frac{64}{1} =64 $
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