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2(x-4)(2x-4)=32
We move all terms to the left:
2(x-4)(2x-4)-(32)=0
We multiply parentheses ..
2(+2x^2-4x-8x+16)-32=0
We multiply parentheses
4x^2-8x-16x+32-32=0
We add all the numbers together, and all the variables
4x^2-24x=0
a = 4; b = -24; c = 0;
Δ = b2-4ac
Δ = -242-4·4·0
Δ = 576
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{576}=24$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-24)-24}{2*4}=\frac{0}{8} =0 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-24)+24}{2*4}=\frac{48}{8} =6 $
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