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16w^2+13w-6=0
a = 16; b = 13; c = -6;
Δ = b2-4ac
Δ = 132-4·16·(-6)
Δ = 553
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{553}}{2*16}=\frac{-13-\sqrt{553}}{32} $$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+\sqrt{553}}{2*16}=\frac{-13+\sqrt{553}}{32} $
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