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=-16A^2+64A
We move all terms to the left:
-(-16A^2+64A)=0
We get rid of parentheses
16A^2-64A=0
a = 16; b = -64; c = 0;
Δ = b2-4ac
Δ = -642-4·16·0
Δ = 4096
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{4096}=64$$A_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-64)-64}{2*16}=\frac{0}{32} =0 $$A_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-64)+64}{2*16}=\frac{128}{32} =4 $
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