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3b^2-3b=
We move all terms to the left:
3b^2-3b-()=0
We add all the numbers together, and all the variables
3b^2-3b=0
a = 3; b = -3; c = 0;
Δ = b2-4ac
Δ = -32-4·3·0
Δ = 9
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{9}=3$$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-3}{2*3}=\frac{0}{6} =0 $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+3}{2*3}=\frac{6}{6} =1 $
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