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12b^2+23b+1=0
a = 12; b = 23; c = +1;
Δ = b2-4ac
Δ = 232-4·12·1
Δ = 481
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}$$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(23)-\sqrt{481}}{2*12}=\frac{-23-\sqrt{481}}{24} $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+\sqrt{481}}{2*12}=\frac{-23+\sqrt{481}}{24} $
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