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(A)=20A-0.5A^2
We move all terms to the left:
(A)-(20A-0.5A^2)=0
We get rid of parentheses
0.5A^2-20A+A=0
We add all the numbers together, and all the variables
0.5A^2-19A=0
a = 0.5; b = -19; c = 0;
Δ = b2-4ac
Δ = -192-4·0.5·0
Δ = 361
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$A_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$A_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{361}=19$$A_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-19}{2*0.5}=\frac{0}{1} =0 $$A_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+19}{2*0.5}=\frac{38}{1} =38 $
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