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64a^2=1
We move all terms to the left:
64a^2-(1)=0
a = 64; b = 0; c = -1;
Δ = b2-4ac
Δ = 02-4·64·(-1)
Δ = 256
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{256}=16$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16}{2*64}=\frac{-16}{128} =-1/8 $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16}{2*64}=\frac{16}{128} =1/8 $
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