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12b^2-27=0
a = 12; b = 0; c = -27;
Δ = b2-4ac
Δ = 02-4·12·(-27)
Δ = 1296
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{1296}=36$$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-36}{2*12}=\frac{-36}{24} =-1+1/2 $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+36}{2*12}=\frac{36}{24} =1+1/2 $
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