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(2x-4)(2x+4)=46
We move all terms to the left:
(2x-4)(2x+4)-(46)=0
We use the square of the difference formula
4x^2-16-46=0
We add all the numbers together, and all the variables
4x^2-62=0
a = 4; b = 0; c = -62;
Δ = b2-4ac
Δ = 02-4·4·(-62)
Δ = 992
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{992}=\sqrt{16*62}=\sqrt{16}*\sqrt{62}=4\sqrt{62}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{62}}{2*4}=\frac{0-4\sqrt{62}}{8} =-\frac{4\sqrt{62}}{8} =-\frac{\sqrt{62}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{62}}{2*4}=\frac{0+4\sqrt{62}}{8} =\frac{4\sqrt{62}}{8} =\frac{\sqrt{62}}{2} $
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