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-3b(4b-1)=0
We multiply parentheses
-12b^2+3b=0
a = -12; b = 3; c = 0;
Δ = b2-4ac
Δ = 32-4·(-12)·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$$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-3}{2*-12}=\frac{-6}{-24} =1/4 $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+3}{2*-12}=\frac{0}{-24} =0 $
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