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33w^2+16w=0
a = 33; b = 16; c = 0;
Δ = b2-4ac
Δ = 162-4·33·0
Δ = 256
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{256}=16$$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-16}{2*33}=\frac{-32}{66} =-16/33 $$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+16}{2*33}=\frac{0}{66} =0 $
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