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12b^2-15b+12b=0
We add all the numbers together, and all the variables
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$$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-3}{2*12}=\frac{0}{24} =0 $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+3}{2*12}=\frac{6}{24} =1/4 $
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