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u2=36u
We move all terms to the left:
u2-(36u)=0
We add all the numbers together, and all the variables
u^2-36u=0
a = 1; b = -36; c = 0;
Δ = b2-4ac
Δ = -362-4·1·0
Δ = 1296
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{1296}=36$$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-36)-36}{2*1}=\frac{0}{2} =0 $$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-36)+36}{2*1}=\frac{72}{2} =36 $
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