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4c(c-2)=0
We multiply parentheses
4c^2-8c=0
a = 4; b = -8; c = 0;
Δ = b2-4ac
Δ = -82-4·4·0
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{64}=8$$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-8}{2*4}=\frac{0}{8} =0 $$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+8}{2*4}=\frac{16}{8} =2 $
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