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4x(2x-4)=8
We move all terms to the left:
4x(2x-4)-(8)=0
We multiply parentheses
8x^2-16x-8=0
a = 8; b = -16; c = -8;
Δ = b2-4ac
Δ = -162-4·8·(-8)
Δ = 512
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{512}=\sqrt{256*2}=\sqrt{256}*\sqrt{2}=16\sqrt{2}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-16\sqrt{2}}{2*8}=\frac{16-16\sqrt{2}}{16} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+16\sqrt{2}}{2*8}=\frac{16+16\sqrt{2}}{16} $
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