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16w^2+31w=0
a = 16; b = 31; c = 0;
Δ = b2-4ac
Δ = 312-4·16·0
Δ = 961
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}$$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{961}=31$$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(31)-31}{2*16}=\frac{-62}{32} =-1+15/16 $$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(31)+31}{2*16}=\frac{0}{32} =0 $
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