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(2x)^2=1024
We move all terms to the left:
(2x)^2-(1024)=0
a = 2; b = 0; c = -1024;
Δ = b2-4ac
Δ = 02-4·2·(-1024)
Δ = 8192
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{8192}=\sqrt{4096*2}=\sqrt{4096}*\sqrt{2}=64\sqrt{2}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-64\sqrt{2}}{2*2}=\frac{0-64\sqrt{2}}{4} =-\frac{64\sqrt{2}}{4} =-16\sqrt{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+64\sqrt{2}}{2*2}=\frac{0+64\sqrt{2}}{4} =\frac{64\sqrt{2}}{4} =16\sqrt{2} $
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