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4u^2=4u
We move all terms to the left:
4u^2-(4u)=0
a = 4; b = -4; c = 0;
Δ = b2-4ac
Δ = -42-4·4·0
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{16}=4$$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4}{2*4}=\frac{0}{8} =0 $$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4}{2*4}=\frac{8}{8} =1 $
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