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12b^2+3b-8=0
a = 12; b = 3; c = -8;
Δ = b2-4ac
Δ = 32-4·12·(-8)
Δ = 393
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{-(3)-\sqrt{393}}{2*12}=\frac{-3-\sqrt{393}}{24} $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{393}}{2*12}=\frac{-3+\sqrt{393}}{24} $
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