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(4x-16)(2x-4)=0
We multiply parentheses ..
(+8x^2-16x-32x+64)=0
We get rid of parentheses
8x^2-16x-32x+64=0
We add all the numbers together, and all the variables
8x^2-48x+64=0
a = 8; b = -48; c = +64;
Δ = b2-4ac
Δ = -482-4·8·64
Δ = 256
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{256}=16$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-48)-16}{2*8}=\frac{32}{16} =2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-48)+16}{2*8}=\frac{64}{16} =4 $
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