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3b(b-2)=12b
We move all terms to the left:
3b(b-2)-(12b)=0
We add all the numbers together, and all the variables
-12b+3b(b-2)=0
We multiply parentheses
3b^2-12b-6b=0
We add all the numbers together, and all the variables
3b^2-18b=0
a = 3; b = -18; c = 0;
Δ = b2-4ac
Δ = -182-4·3·0
Δ = 324
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{324}=18$$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-18}{2*3}=\frac{0}{6} =0 $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+18}{2*3}=\frac{36}{6} =6 $
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