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b(b=12)=0
We move all terms to the left:
b(b-(12))=0
We multiply parentheses
b^2-12b=0
a = 1; b = -12; c = 0;
Δ = b2-4ac
Δ = -122-4·1·0
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{144}=12$$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-12}{2*1}=\frac{0}{2} =0 $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+12}{2*1}=\frac{24}{2} =12 $
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