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12w^2-13w-2=0
a = 12; b = -13; c = -2;
Δ = b2-4ac
Δ = -132-4·12·(-2)
Δ = 265
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-13)-\sqrt{265}}{2*12}=\frac{13-\sqrt{265}}{24} $$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+\sqrt{265}}{2*12}=\frac{13+\sqrt{265}}{24} $
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