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=-A^2+10A
We move all terms to the left:
-(-A^2+10A)=0
We get rid of parentheses
A^2-10A=0
a = 1; b = -10; c = 0;
Δ = b2-4ac
Δ = -102-4·1·0
Δ = 100
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{100}=10$$A_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-10}{2*1}=\frac{0}{2} =0 $$A_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+10}{2*1}=\frac{20}{2} =10 $
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